Блог пользователя djm03178

Автор djm03178, история, 5 лет назад, По-английски

안녕하세요, 코드포스! (Hello, Codeforces!)

I'm glad to invite you to Codeforces Round 620 (Div. 2). The contest will start at Feb/15/2020 16:05 (Moscow time), and it is rated for all participants with ratings under 2100.

You will be given 6 problems and one of the problems has 2 subtasks. The contest duration is 2 hours. The score distribution will be announced later.

All problems are prepared by me, with huge help from the testers with developing great solutions.

I'll be on the community Discord server after the contest to share my thoughts and get feedback about the problems.

Thanks to 79brue, molamola., Savior-of-Cross, evenharder, cs71107, Justice_Hui, rkm0959, chpark1111, imeimi, InfiniteChallenge, gaelim, jh05013 (Good tester), yuto0518, N_jara, aryanc403, SnowGoose, solvemproblr, surung9898, and ko_osaga for testing the round. I would also like to thank 300iq for round coordination, and MikeMirzayanov for the great Codeforces and Polygon system.

Hope you enjoy the problems!

UPD: The scoring distribution is 500 — 1000 — 1500 — 1750 — 2000 — (2000 + 1000)

UPD2: The contest is finished! Thanks so much for your participation! The editorial is here.

UPD3: Congratulations to the winners!

Div. 2

1: Itst_boyfriend

2: COVID-19

3: naive_xay

4: anta.baka

5: ChitandaEru

Unofficial Div. 1

1: wucstdio

2: ksun48

3: jiangly

4: uwi

5: teapotd

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5 лет назад, # |
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Wow! Another Korean Round! I hope you enjoy this round. I am a tester of this round, and I think you would enjoy! Good luck and have fun.

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5 лет назад, # |
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I don't think jh05013 is a real newbie.If you see in his first two contests he solved 4 problems and then i think he just didn't care about his profile maybe because he likes being a tester more , so he just intentionally solved 0 problems.And no offence jh05013 but you are just 3 points of rating away,breaking the world record of the most newbie.

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    5 лет назад, # ^ |
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    he said in the linked thing that contests were too stressful for him (he was 1700+). Now he just don't care to mess up :)

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      5 лет назад, # ^ |
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      Moreover, in this link, he said he is willing to inspect contests(though these are other site contest), and said to call him a tester if someone wants to. Based on the positive response people have to this article I can conclude he is a good tester.

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5 лет назад, # |
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Finally a newbie Tester this time... So all colors tested this round...sweet

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5 лет назад, # |
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우왕!

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5 лет назад, # |
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colorful testers

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5 лет назад, # |
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Fan E AE YO

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5 лет назад, # |
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:v Testers in every ranks !

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5 лет назад, # |
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G

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    5 лет назад, # ^ |
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    Alright the time will be changed accordingly as cf revolves around your schedule :p (not rlly)

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5 лет назад, # |
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Editorials ?????

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5 лет назад, # |
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Hope B will be clear this time. Past 2 contests Test Cases for B were very hard.

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5 лет назад, # |
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OwO,well...Korea round is my nightmare,hope pass C

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5 лет назад, # |
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As one of the lower ranked testers, I would highly recommend participation in this round. These are good problems, there is something for every skill level here!

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5 лет назад, # |
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I wish positive rating for every participant :)

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    5 лет назад, # ^ |
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    We all wish that but this is not possible.

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    5 лет назад, # ^ |
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    The total rating decreases after every contest, otherwise, ratings will become higher and higher because new users will join Codeforces with the initial 1500 rating points.

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5 лет назад, # |
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Round with subtask

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팬이에요! :fan:

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5 лет назад, # |
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Can someone tell me what does one of the problems has 2 subtasks means? thanks in advance.

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    5 лет назад, # ^ |
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    I'm not quite sure, but it may be an easy version and a hard version of the same problem. Like C1 & C2 for example.

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    5 лет назад, # ^ |
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    One of the problems has an easy version and a hard version. You'll know which problem has subtasks after the contest starts.

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5 лет назад, # |
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I hope the problem statements are short and clear.

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5 лет назад, # |
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Want to be Pupil today.Wish me good luck

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5 лет назад, # |
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I wonder what is jh05013's real rating.

Because when he became blue.I can only solve div.2B problems

Maybe he have a real rating of Grandmaster or higher :>

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5 лет назад, # |
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I found a really interesting thing that jh05013 and I have both the lowest rating -43 and after that, our ratings both increase 3.

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5 лет назад, # |
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How to solve E?

Edited : Thank you to all who replied. Because I got TLE on pretest 14 during the contest, I'll try again using the faster LCA algorithm.

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    5 лет назад, # ^ |
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    Hint: The added edge is there for parity reasons.

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    5 лет назад, # ^ |
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    only three possible ways l = a -> x -> y -> b l = a -> y -> x -> b l = a -> b

    if l<=k and l%2==k%2 then we have a solution

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      5 лет назад, # ^ |
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      Did the same but didn't work :(

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      5 лет назад, # ^ |
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      I do not get it how to find the shortest possible path after adding x,y, we need that to check against k. Somebody explain?

      Edit: Or is it expected to do a BFS to find it? That would be to slow.

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        5 лет назад, # ^ |
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        You need to make use of precomputing depths, and LCA ( Least Common Ancestor ) of two nodes. Read more here.

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          5 лет назад, # ^ |
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          The LCA does not help if the path after adding the new edge x,y is shorter than the one through the LCA.

          Edit: After reading the other answer below I understand. We simply use LCA again to find path length from a to x and y etc. Thanks for explanation.

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            5 лет назад, # ^ |
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            Read my comment below. After adding the edge the shortest path will have to be one of the three alternatives. ( It can be proven ).

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    5 лет назад, # ^ |
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    The path between two vertices in the query may either go through the added edge or not. Analyze these cases individually, finding path lengths in O(1) using whatever fast LCA algorithm you prefer. Then if the difference between the length of the path obtained in some of these ways and the desired length if divisible by 2, answer is yes, otherwise no.

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    5 лет назад, # ^ |
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    Root the tree at an arbitrary vertex. Precompute depths of each node to give distances between two nodes in $$$O(log(N))$$$ time using LCA.

    When adding a new edge $$$(x,y)$$$, there are now three possible paths ( without going back and forth ), which are

    1. original path between $$$(a,b)$$$

    2. $$$(a,x), (x,y), (y,b)$$$

    3. $$$(a,y), (y,x), (x,b)$$$

    Now, if any of these paths can reach $$$b$$$ from $$$a$$$ in less than or equal to $$$k$$$ steps and has same parity as $$$k$$$, then answer is YES.

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    5 лет назад, # ^ |
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    First we do not consider added edge. We can 'hang out' instead of actually making meaningful move as much as we need — which is, if we can arrive $$$b$$$ at $$$t$$$, we can hang out some even number of steps loafing back and forth.

    Then we consider new edge. Since new edge form a cycle, it opens new possibility — by loafing around in cycle. If cycle length is $$$l$$$, we can arrive at

    $$$p + a\times l + 2b$$$

    for any arbitrary nonnegative integer $a, b$, where $$$p$$$ is shortest time we can arrive at $$$b$$$ using the added edge. (If we are not using the new edge, we can't hang out on the cycle).

    All we need is careful casework (there aren't many) and computing distance in trees using LCA or some other fast methods.

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5 лет назад, # |
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Am I the only one who solved A with binary search instead of much more obvious 1 liner solution?

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    5 лет назад, # ^ |
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    If there is the point to which we get after $$$k$$$ steps from both $$$x$$$ and $$$y$$$ is

    $$$x + k \cdot a = y - k \cdot b$$$ $$$k \cdot a + k \cdot b = y - x$$$ $$$k\cdot (a+b) = y -x$$$ $$$k = \frac{y-x}{a+b}$$$

    So, if $$$(y-x)\%(a+b)=0$$$

    then answer is $$$k$$$, in other case the answer is $$$-1$$$.

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    5 лет назад, # ^ |
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    I used binary search too, and needed the editorial to make me think of manipulating the equation :)

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5 лет назад, # |
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How to solve C?

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    5 лет назад, # ^ |
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    After each customer comes in, you have an effective range of temperature, let it be (x, y). For the next customer find the intersection of (x — t, y + t) (where t is difference between entry times) and his temperature range. That is your new effective range. Initial range is (m, m). If all the intersections are non-empty you can satisfy, otherwise no.

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    5 лет назад, # ^ |
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    • Group all customers attending at a given time t.
    • For each group check if there is an overlapping temperature range and store that. For instance if the ranges are [(1, 4), (2, 5)], the overlapping range is (2, 4).
    • If it's not possible to form an overlapping range at any t, then it's not possible
    • Assuming each t has a range, check if it's possible to move from range at ti to range at ti+1. Also update the range of temperatures possible at each step.
    • If possible range at step i is (lo, hi), Then possible range at step i + 1 is (lo - (ti+1 - ti), hi + (ti + 1 - ti))
    • Just check if that is acceptable for all customers at ti+1 and update lo and hi. (Not sure if this step can be optimised further but it passes the problem limits)

    My submission 71152069

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    5 лет назад, # ^ |
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    I have a solution that is easier to implement in my opinion.

    Loop through all pairs of customers. If it is impossible to reach the second Customer's range from the 1st Customer's range in the time difference between them, then the answer is impossible.

    O(N^2) solution.

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5 лет назад, # |
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How to solve C and D guys ?

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    5 лет назад, # ^ |
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    For C

    Let's say that we have minimal and maximal temperatures with which we can get to the next consumer, and if the intersection of our range and range (l,r) is not empty, then it's okay, now we just update our range — basically it would be equal to intersection of ours and (l, r).

    My submission: 71146715

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    5 лет назад, # ^ |
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    For D

    The main idea here is that for minimum we should build our array in such a way that bigger numbers have lower indexes and for maximum bigger numbers should have bigger indexes.

    You can chek out my submission: 71159214

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5 лет назад, # |
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how to solve D?

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    5 лет назад, # ^ |
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    For Min: starting from the first '<', start at n and go down but if there are <<<< or something sort that subarray then for the rest just go down by 1 each time so say for the 2nd sample, my answer would be 5 4 3 7 2 1 6

    Max: for max it's the same thing but all the numbers after a < are going up so 5 4 3 6 2 1 7

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5 лет назад, # |
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Thanks for the contest, have a good day problem setters <3

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5 лет назад, # |
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In Problem B of DIV2. the problem statement mentioned that the string can be "REORDERED" and not "REVERSED" but my code was failing the first testcase here : https://codeforces.me/contest/1304/submission/71145055

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5 лет назад, # |
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Can someone please tell me why this code for C is giving rte. The only place where an exception can be thrown is the assert in which case the test cases are wrong. https://codeforces.me/contest/1304/submission/71155599

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    5 лет назад, # ^ |
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    Perhaps it has to do with your usage of the word "time"? I think it's a reserved keyword that can still do arithmetic with integers. You're right, the code itself doesn't seem like it can segfault as far as I can see.

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      5 лет назад, # ^ |
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      I had submitted the same code without the assert line. That gave wa.

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        5 лет назад, # ^ |
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        That doesn't necessarily imply much; undefined behavior is still within your set of possibilities. In fact, I'm much more inclined to think that you have UB somewhere; this would probably cause a segfault sometimes and a WA at others.

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          5 лет назад, # ^ |
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          The thing is that I have been getting RTE only when I have included assert in the code. I have submitted the same code without assert thrice (introducing one useless variable each time so that it allows me to submit) and got WA in all those.

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            5 лет назад, # ^ |
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            I've gotten several questions regarding this.

            The problem is that you're "returning" the function while you haven't gotten the full test case yet. Some people used 'break' inside a loop. These codes generally got incorrect on pretest 3.

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              5 лет назад, # ^ |
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              Yes that's the mistake. Thanks for pointing out and sorry for wasting your time.

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    5 лет назад, # ^ |
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    I got similar Runtime error in Problem B today. It was resolved by removing #define int long long. I couldn't figure out Why it happened today, I haven't faced any problem due to this statement earlier.

    My Submission link: Submission

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    5 лет назад, # ^ |
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    You are returning from function, "solveTestCase", before taking all inputs. Instead return only at the end of the function.

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5 лет назад, # |
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Every time........!

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5 лет назад, # |
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Thanks for the round, hope pretests were good enough.... I wish everyone to increase their rating ;)

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5 лет назад, # |
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Oh,just five minutes after the contest over my solution passed all the examples.:( The code for the round are a little bit long. I think a longer time will be better.:)

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5 лет назад, # |
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Why does codeforces skip main test and don't take the best submission? I missed my chance of becoming Master today.

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5 лет назад, # |
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I don't understand what is wrong in my submission, the answer is printed but still it shows run time error. Here is my submission 71154644

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5 лет назад, # |
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here i come #808080

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5 лет назад, # |
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really sad, failed B on test 13.

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5 лет назад, # |
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It was such a good contest for the beginners. As the problems were not so hard as before :) But I think it should be harder, especially the Problem F.

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5 лет назад, # |
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My color is going to change. I love these questions!!!

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5 лет назад, # |
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Thank you for strong pretests!

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5 лет назад, # |
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Question B : Why there is run time error coming in https://codeforces.me/contest/1304/submission/71161367 this code can anyone tell me?

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5 лет назад, # |
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5 лет назад, # |
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What is the mistake in this code, I am getting wa on prestest 8

code: https://codeforces.me/contest/1304/submission/71165457

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    5 лет назад, # ^ |
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    If you erase elements in a set while iterating through it, especially your current pointer, you might end up skipping a few elements. This is shown when you run your code on the following case:

    4 2
    ab
    cd
    ba
    dc
    

    A different approach is to scrap the set entirely and instead use an array of strings, with boolean markers that say if you've used that string already (example, with the mk array).

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      5 лет назад, # ^ |
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      Thanks a lot man!

      But why does it skips a few elements?

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        5 лет назад, # ^ |
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        I didn't realize that you called a vector s, but it functions the same as a set in this context. My bad.

        Imagine your vector like this:

        ab <- (pointer)
        cd
        ba
        dc
        

        Now you find ba to be a match with ab, so you erase them both. Where does the pointer go? It can only advance to cd:

        cd <-
        dc
        

        Now you increment the pointer, and it ends up at:

        cd
        dc <-
        

        Since you only check ahead of your current iterator for matches, you don't find anything else and terminate. There's probably a topic on StackOverflow addressing this issue, but the basic premise is to only increment when you don't erase.

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5 лет назад, # |
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I believe test cases for D are a bit weak?I changed my code for D a bit during contest as it had 1 mistake, but after the contest, it still passed....

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5 лет назад, # |
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71166353 works on pretest 4 on my machine, and on ideone link but failed during contest and also on custom invocation. MikeMirzayanov djm03178

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5 лет назад, # |
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Извините, за что анрейт, если у меня списали задачу?

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5 лет назад, # |
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My solution of B is failing in this test case but somehow it managed to pass system testing —
3 3
tab
bat
tab

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5 лет назад, # |
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where i am wrong? https://codeforces.me/contest/1304/submission/71174871 problem C

i am iterating in reverse, finding the largest intersection possible so that both current and previous customer are satisfied with temperature

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5 лет назад, # |
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Для решения задачи E использовал код для нахождения LCA из этого этого источника. Пришло письмо о совпадении кода с другими участниками и нарушении правил. Прошу восстановить мои результаты в контесте, так как использование публичных источников не запрещено.

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5 лет назад, # |
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Not sure if I should be posting this here, but:

I just got a message saying that my solution for problem E coincides with that of some other participants. This is likely because I used the CP Algorithms implementation for LCA which you can find here: https://cp-algorithms.com/graph/lca.html

Although I added a few additional methods in my source code, a big chunk of my solution was for the LCA structure. So I guess that could've set off the response. I hope this clears things up.

If you need any more information, please let me know. MikeMirzayanov

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5 лет назад, # |
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I am not sure whether I should comment here . but i just got a warning about coinciding my code with another user.Since I am new here , I dint know about the codes copying from ideone . I was using IDEONE in default setting. My code is https://codeforces.me/contest/1304/submission/71137281. And the guy who copied my code is https://codeforces.me/profile/8bit_thug I request you not to do this. this kills contest spirit. And yeah Since He submmitted code after 3 minutes when I submitted hence it is cleared that he copied my code. Moreover all his submissions are skipped. Do I need to provide more information? MikeMirzayanov

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5 лет назад, # |
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My solutions for this contest were skipped most likely because I used the implementation of LCA from here which is probably where some other participants picked the code up from.

djm03178 can you look into this?

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5 лет назад, # |
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Мое решение 71159863 по задаче 1304E совпадает с решением другого пользователя, так как видимо мы использовали один и тот же источник, а именно https://e-maxx.ru/algo/lca.

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5 лет назад, # |
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I got a message from User system regarding code similarity about the E problem (71164831). I have used the code for lowest common ancestor from this site (same code published in the site mentioned by other affected users) which was last updated on 15th June 2018 (here) which is conclusive evidence of this being published well before the contest.

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5 лет назад, # |
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I hope this is the right place to put this, otherwise, please tell me where I should write this. I received the following message:

Attention!

Your solution 71159069 for the problem 1304E significantly coincides with solutions ksun48/71134293, hoke_t/71152328, Venia/71156816, hanga97/71159069. Such a coincidence is a clear rules violation. Note that unintentional leakage is also a violation. For example, do not use ideone.com with the default settings (public access to your code). If you have conclusive evidence that a coincidence has occurred due to the use of a common source published before the competition, write a comment to post about the round with all the details. More information can be found at http://codeforces.me/blog/entry/8790. Such violation of the rules may be the reason for blocking your account or other penalties. In case of repeated violations, your account may be blocked.

For the solution of problem E, I used the RMQ (range minimum query) and LCA (lowest common ancestor) implementations from https://github.com/kth-competitive-programming/kactl/

Specifically from https://github.com/kth-competitive-programming/kactl/blob/master/content/data-structures/RMQ.h and https://github.com/kth-competitive-programming/kactl/blob/master/content/graph/LCA.h

This code was online long before the start of the round, so I hope, this issue can be resolved.

Thanks for the round btw. I enjoyed it.

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5 лет назад, # |
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Can someone answer why the compiler of Codefoces gives run time error if i have not type casted size of a vector whereas most of the other compilers don't give that error. In this contest only i got stuck in problem B due to this error.

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    5 лет назад, # ^ |
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    Me too

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    5 лет назад, # ^ |
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    the .size() function returns an unsigned integer, so when you try to do.size() — 1 when the size is 0, it overflows and causes runtime error.

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      5 лет назад, # ^ |
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      Got that...but when i did this for(long long i=a.size()-1;i>=0;i--), it gave me run time error but for this for(int i=a.size()-1;i>=0;i--) it did not. Can you tell me why this happened? And thanks in advance :)

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5 лет назад, # |
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Can someone please explain how I am getting a different output for E on my local compiler than the codeforces judge.71178404

I am the getting the expected output on my machine. I believe the logic is correct. https://ideone.com/LfR13i Expected output with same input

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5 лет назад, # |
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Nice problemset!

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5 лет назад, # |
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Hello MikeMirzayanov, I have received a message about being disqualified from the contest, because of plagiarism. I have used the LCA calculation from cp-algorithm. Here is a link to the web archive, as evidence, that the code has been published before the contest. I have not used any online IDE. Kindly look into this, Thanks a lot!

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5 лет назад, # |
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Hey,
My solutions were skipped because one of them coincided with solutions of other participants.
It was 100% caused because of this implementation of LCA, which I use for a long time.
Regarding rules, one can use publicly available implementations posted before the round.
MikeMirzayanov can you do something about this?

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5 лет назад, # |
Rev. 2   Проголосовать: нравится +29 Проголосовать: не нравится

Me in this contest:

Firstly, solve problem C, and then B-A-D-E-F.

After solving all the problems, I looked at the standings, and found rank 2 is only 61 points lower than me, and he has locked all his problems and started hacking.

So I locked my ABCDE and started hacking too.

When I opened a solution to problem B, I was surprised to find there was a very silly BUG. A code with such silly BUG wouldn't pass the pretests.

So I took a look at the statement, only to find that I had misread the problem. I thought I can reverse any strings as well as reorder them.

1000 points! If I had not made such a silly mistake, I would probably get on the 1st place, but now I couldn't resubmit because I had locked my problem. So I poured out my sadness to jiangly, who had solved all the problems too and become rank 3 at that time.

Here is the chat history:

Yes, I misread the problem again. I forgot that all the strings are distinct.

So from my story we can conclude that it's very important to read the statement carefully, but if you misread the statement, you still have chance to pass the problem.:)

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    5 лет назад, # ^ |
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    Lol same. I also thought I need to check the length of the only palindromic string I will add in the middle and changed my code in the last minute of the contest. And after 5 minutes comes the realization that all the string have the same length. Lost 500 points and went from +45 to -13.

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    5 лет назад, # ^ |
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    I just found out that I also misread. Thank you.

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5 лет назад, # |
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In problem D, you can solve for max LIS and re-use it to find min LIS by manipulating the string, ie. reversing and inversing — replacing $$$<$$$ and $$$>$$$ with one another. Here is my solution.

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5 лет назад, # |
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Great contest!

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5 лет назад, # |
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My contest history for this contest is suddenly deleted.Why this happened?

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5 лет назад, # |
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apology for poor english

where were you when top rate dies

i was sat at work writin code when telegram ring

'top rated is die'

'no'

and you???????

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5 лет назад, # |
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hello in solution B try this 2 3 AAA AAA and it will answer you 3 AAA i think it should be 6 AAAAAA anyone can help me with this i can't see in description that say the same string doesn't count

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4 года назад, # |
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great contest, but I would have liked if you could have removed problem F1. It is extremely trivial (I think div2C). F2 is the real deal and is a great problem. There was no reason to add F1. (and just because of the long legend and its place in the problemset, F1 is rated 2300, while it is more like 1500).

Again, great problems, especially E and F2, but don't add subtasks which add no value to the problemset.

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    4 года назад, # ^ |
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    I agree that F1 isn't very interesting by itself, but the subtask is there mostly because of the difficulty gap between E and F2, and I don't think it's that trivial to be a D2C as you can see how many participants have solved it (both during contest and post contest). I still think it's at least as hard as E. In fact, F1 was initially planned to be F itself, and I improved the solution to make a harder version which is F2 now.

    In the end, I think it did quite a job being a bridge between E and F2. For those who couldn't generalize the idea for F, they could write a naive solution to try F1 only, and they were rewarded more than those who couldn't even manage to do it in time.

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      4 года назад, # ^ |
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      Well sir, F1 was a very obvious DP, was it not? Once you see the pictures in the samples, you immediately come up with the DP!!!! The transition is so easy! It needs no insight!

      The prefix and suffix maximum precomputation was very natural too, like you first came up with the transition, and then decided "yeah, I need prefix and suffix max".

      As for bridging the difficulty is concerned, on the other hand, I feel this problem left div2 participants with an even less amount of time for F2. F2 was kinda short to write but because of lazy segtree (in my solution) it is potentially hard to debug. A participant may feel:

      F1 is so trivial that I might as well finish that problem first as it is worth 2000 points

      but of course you don't agree that the DP was trivial. But I showed this problem to my friends and they agreed that F1 was instantly obvious. F2 on the other hand I really enjoyed, thank you.