vovuh's blog

By vovuh, history, 3 weeks ago, translation, In English,

UPD: Pay attention to the changed start time of the competition.

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Hello! Codeforces Round #627 (Div. 3) will start at Mar/12/2020 16:05 (Moscow time). You will be offered 6 or 7 problems (or 8) with expected difficulties to compose an interesting competition for participants with ratings up to 1600. However, all of you who wish to take part and have rating 1600 or higher, can register for the round unofficially.

The round will be hosted by rules of educational rounds (extended ACM-ICPC). Thus, during the round, solutions will be judged on preliminary tests, and after the round it will be a 12-hour phase of open hacks. I tried to make strong tests — just like you will be upset if many solutions fail after the contest is over.

You will be given 6 or 7 (or 8) problems and 2 hours to solve them.

Note that the penalty for the wrong submission in this round (and the following Div. 3 rounds) is 10 minutes.

Remember that only the trusted participants of the third division will be included in the official standings table. As it is written by link, this is a compulsory measure for combating unsporting behavior. To qualify as a trusted participants of the third division, you must:

  • take part in at least two rated rounds (and solve at least one problem in each of them),
  • do not have a point of 1900 or higher in the rating.

Regardless of whether you are a trusted participant of the third division or not, if your rating is less than 1600, then the round will be rated for you.

Thanks to MikeMirzayanov for the platform, help with ideas for problems and for coordination of my work. Thanks to my good friends Daria ZeroAmbition Stepanova, Mikhail pikmike Piklyaev, Maksim Ne0n25 Mescheryakov and Ivan BledDest Androsov for help in round preparation and testing the round.

Good luck!

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Thanks to Artem Rox Plotkin and Dmitrii _overrated_ Umnov for help with testing the round!

UPD2: Editorial is published!

 
 
 
 
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3 weeks ago, # |
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"You will be given 6 or 7 (or 8) problems and 2 hours to solve them."

Does this means that the problems haven't been decided ?

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    3 weeks ago, # ^ |
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    No, Because it is almost-copy-pasted-part

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    3 weeks ago, # ^ |
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    The numbers of questions has been decides, maybe they will have a little change before contest.

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    3 weeks ago, # ^ |
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    May not. I think it's just a general idea so that almost every contest could use this template. ^-^

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Which contest coincided with the original timing?

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Good luck everyone !!!

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Who else is hyped for Div 3 because can't get good rating from Div 2?

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What does mean by "almost-copy-pasted-part"?

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    3 weeks ago, # ^ |
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    The "almost-copy-pasted-part" is almost equal in the announcements of most contests

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So excited for first contest. Good luck everybody!

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How was your Holi celebration ? Was anyone not able to play due to coronavirus threat?

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    3 weeks ago, # ^ |
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    So many downvotes :( . I haven't asked anything wrong :( .

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2 hours in Div 3 and 4 hours in Reply Code Challenge, it's going to be an exhaustive day.

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Hope that I can solve A-B-C.

Good luck guys <3

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3 weeks ago, # |
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well,is there anyone know what's the name of the other contest(1.5hours later) after this contest?

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3 weeks ago, # |
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Hope to become expert after this round

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3 weeks ago, # |
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I hope to get blue in this contest

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3 weeks ago, # |
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I don't think i can solve over two problems...

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    3 weeks ago, # ^ |
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    if you think you can't, so you can't.

    Believing in yourself is the first step to success.

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    3 weeks ago, # ^ |
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    you were wrong

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3 weeks ago, # |
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Hope problems will not be indirect like this

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    3 weeks ago, # ^ |
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    Well, you can jump through the centre.

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      3 weeks ago, # ^ |
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      Ya I can for that I require wings of Advanced mathematics and data structures. Anyways Good Luck to everyone for the contest.

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3 weeks ago, # |
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Bad luck for me becoz i can't attend this contest.

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3 weeks ago, # |
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Good luck Everyone

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3 weeks ago, # |
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Hope i will be green after this round :))

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3 weeks ago, # |
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I use this site during contest :)

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3 weeks ago, # |
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I got logged out 2-3 times during contest. Did it happen to anybody else or problem at my end?

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    3 weeks ago, # ^ |
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    Same here :(

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    3 weeks ago, # ^ |
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    Glad to see I wasn't the only one who faced the same issue.
    Made me wonder if my account got hacked or something :/

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    3 weeks ago, # ^ |
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    Well, the thing you had also happened to me, so I felt unhappy during the contest:(

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    3 weeks ago, # ^ |
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    Yes,this happened with me also. Maybe a coincidence, but it happened mostly when I clicked submit button.

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    3 weeks ago, # ^ |
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    Me too. I thought it was my browser problem or something like that and erased cache and it happend again... :(

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    3 weeks ago, # ^ |
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    same here, maybe it's the browser's problem or something else, idk

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    3 weeks ago, # ^ |
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    Same to me ((

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3 weeks ago, # |
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I feel it difficult to understand some of the problem statements,does anyone feel the same?

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  • A Done
  • B Done
  • C Done
  • Me Done
  • Nice Contest
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3 weeks ago, # |
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Good to see that vovuh is the writer.

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3 weeks ago, # |
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The problems are too easy so that more than 500 participants solved all the problems

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Guys,I have a problem! do not have a point of 1900 or higher in the rating Is that means standings in contest is different from your rank which is the final standings?

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    3 weeks ago, # ^ |
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    The contest will be rated for those participants who have a rating below 1600. However, participants with rating greater than or equal to 1600 can participate unofficially. Separate standings are available for official and unofficial+official participants.

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During the contest, I automatically got logged out from codeforces. I had two tabs of codeforces open in firefox. When I tried to submit I got to know I am not logged in. This has happened thrice. Is it normal/minor bug or there is some other problem ?

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For those who cannot understand what is subgraph and subtree. Click :(

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Thank you very much for the rating friendly contest

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How to solve D ;-;

My approach
My source code

Found Mistake: (wrong formula) cout << 1LL * n * (n + 1) / 2; -> cout << 1LL * n * (n - 1) / 2; Found Mistake: (integer overflow) ll res = q * (q - 1) / 2 + q * zero; -> ll res = 1LL * q * (q - 1) / 2 + 1LL * q * zero;

AC Code (Spoiler warning)
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    3 weeks ago, # ^ |
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    You can use binary search. create a diff array such that diff[i] = teacher[i]-student[i]. sort this array. at a particular point i calculate how much you'll need to compensate for when you choose j. e.g if teacher[i]-student[i] = -2. then we know we need diff[j] >= 3, so as to make teacher[i]+teacher[j] > student[i] + student[j].

    you can easily binary search this value in diff array.

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    3 weeks ago, # ^ |
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    Let $$$c_i = a_i - b_i$$$. Now you have to count the number of pairs of $$$c_i$$$ with the positive sum. It can be done in $$$O(n log n)$$$ with binary search.

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    3 weeks ago, # ^ |
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    Create an array of differences of a[i]-b[i] then use lower bound to find good pairs for negative numbers and zero numbers and for positive use binomial coefficient (ncr).

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    3 weeks ago, # ^ |
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    I used PBDS.

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      3 weeks ago, # ^ |
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      how?

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        3 weeks ago, # ^ |
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        Let's just iterate over the array from the start, and since we need to find unordered pairs, so for a particular i, we need all elements before it such that: ai — bi > bj — aj or Let's denote bj — aj as val, then we need all such elements where : val < ai — bi So PBDS on pair of elements acts like a multiset with additional feature ,giving count of elements less than x too.So use that function for each i and get count of elements less than {ai-bi-1,inf}, and add it to answer and insert pair {bi-ai,i} to the pbds.

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          3 weeks ago, # ^ |
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          Can u please explain — "st.order_of_key(mp(arr[i]-arr1[i]-1,inf))"--- In this part of your code,why you used inf?

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        3 weeks ago, # ^ |
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        Use PBDS as a multiset (comparator less_equal). Traverse from the end of the array and for each index, increment answer by order_of_key of $$$a[i]-b[i]$$$ (basically index of upper bound). Then insert $$$b[i]-a[i]$$$ in the PBDS.

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          3 weeks ago, # ^ |
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          Yes, yes, yes, shooting pigeons with a cannon — that's my style! 73043201

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      3 weeks ago, # ^ |
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      Can you share your code? I tried using pbds, but it gave wa on test 6.

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        3 weeks ago, # ^ |
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        You can look it up in my submissions and if are not permitted, then add me as a friend and then look it up in standings submission. Sharing code here will scribble the comment section.

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    3 weeks ago, # ^ |
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    It can also be done with sorting; if you have a sorted array $$$c$$$ and you want to know for particular $$$i$$$, how many $$$j$$$ satisfy $$$c[i]+c[j]>0$$$, then you can decrement $$$j$$$ from $$$n$$$ down to the first index where $$$c[i]-c[j]<=0$$$, and count $$$n-j$$$. Then if you iterate over increasing $$$i$$$, this $$$j$$$ can only decrease.

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    3 weeks ago, # ^ |
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    What is wrong with my approach ? ;-;

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    3 weeks ago, # ^ |
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    problem is similar to inversion count, so used BIT tree to solve it, after compression.

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      3 weeks ago, # ^ |
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      I'm glad i'm not the only one who used BIT :)))))

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      3 weeks ago, # ^ |
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      I to used BIT to solve this. But I complicated it as hell. ACed

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    3 weeks ago, # ^ |
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    Let c[i] = a[i] — b[i]. Now we have to count the number of pairs with positive sum. Sort array c.

    Set two pointers, one at index 0 (pointer j) and the other at index n-1 (pointer k). In each iteration decrease k by 1 and find the corresponding index j, by running a separate loop within the previous loop. Exit when j >= k.

    Initialize ans = 0, and keep adding (k — j) in each iteration, i.e. ans += k — j

    Note that j keeps on increasing in each iteration. We do not restart j from 0 each time.

    Time Complexity: O(n)

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      3 weeks ago, # ^ |
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      You have to sort array c as well, otherwise this 2 pointer technique will not work.

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    3 weeks ago, # ^ |
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    thanks for taking your time to write an explanation.

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There's no one later than me to solve A! I like playing with the tetris game,but I do not like this problem!

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3 weeks ago, # |
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Can any one give a hack on the following approach for D?

Sort them by $$$a_i-b_i$$$, then use binary search to find last okay one.

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    3 weeks ago, # ^ |
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    Gradually sort?

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    3 weeks ago, # ^ |
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    sorting them will change the order of the indices

    and in the problem you need $$$j$$$ to be $$$>$$$ $$$i$$$

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      3 weeks ago, # ^ |
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      That doesn't matter, if you are counting all distinct pairs $$$(i,j)$$$ which satisfy some property, then this will equal the number of distinct pairs $$$(\sigma(i),\sigma(j))$$$ which satisfy the same property, for any permutation $$$\sigma$$$

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    3 weeks ago, # ^ |
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    deduce condition ,suppose if ai-bi=di; then we need all di+dj>0 (j>i)pairs (counted once)

    for this make a array d=ai-bi you can divide array into three parts — positive(np) ,negative(nn) ,zero(n0); solve separately for zero , positive and negative;

    we can get our condition satisfied in three way , 1. positive di pairs 2. zero paired with any positive di 3. positive negative pairs where magnitude of positive is bigger than negative;

    a1=np(np-1)/2 (Ist cond); a2= n0*np(IInd cond) ; and for a3(IIIrd cond) ,you can loop in (O(nn+np)) time and at last ans=a1+a2+a3;

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    3 weeks ago, # ^ |
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    Make an array of Ai — Bi, sort it and binary search the right one for each. That's the solution I think.

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    3 weeks ago, # ^ |
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    Maybe you can use the Binary Indexed Tree in value.

    First discrete all $$$a_i - b_i$$$ and $$$b_i - a_i$$$.

    Then for every $$$i \in [2, n]$$$, calculate all for $$$j \in [1, i)$$$ which $$$b_j - a_j < a_i - b_i$$$

    The time complexity will be $$$O(n \log n)$$$.

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Lightning-fast system testing

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    3 weeks ago, # ^ |
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    There is a 12 hr hacking phase, after which system testing will begin.

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      Codeforces has large servers which can carry out fast testing.

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What is the 8 test case for problem D?

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Why, i kept logging out after few intervals ,throughout the contest??

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How to solve E? Also, what is the name of the concept used in E?

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    3 weeks ago, # ^ |
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    I believe it's called dynamic programming.

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    3 weeks ago, # ^ |
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    Dynamic programming. $$$dp_{i, j}$$$ is the number of good times when we consider the first $$$i$$$ times ($$$0$$$-indexed) and did this $$$-1$$$ "operation" $$$j$$$ times. The answer is $$$\max\limits_{j=0}^{n} dp_n$$$.

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      3 weeks ago, # ^ |
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      what's the meaning of the first $$$i$$$ times,I can't get it

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        3 weeks ago, # ^ |
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        More correctly, $$$dp_{i, j}$$$ is the number of "good times" when we consider first $$$i$$$ values $$$a_i$$$ and used $$$-1$$$ $$$j$$$ times. Transitions are also pretty easy: if the sum of the first $$$i$$$ values $$$a_i$$$ is $$$s$$$ then $$$dp_{i + 1, j} = max(dp_{i + 1, j}, dp_{i, j} + [(s - j) \% h \in [l; r]])$$$ and $$$dp_{i + 1, j + 1} = max(dp_{i + 1, j + 1}, dp_{i, j} + [(s - j - 1) \% h \in [l; r]])$$$. $$$\%$$$ is modulo operation ofc.

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          3 weeks ago, # ^ |
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          I understand that now, thank you very much~~

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    3 weeks ago, # ^ |
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    state are index and modulo

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in problem D , solution of 10^10 was giving TLE for test case 12,althought time limit was 2 seconds, http://codeforces.com/contest/1324/submission/73076815 i thought 10^10 solutions can run in 2 sec

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Downvoted because the sentence in problem E is completely unreadable.

The sentence "The i-th time he will sleep exactly after ai hours from the time he woke up" and "Vova can control himself and before the i-th time can choose between two options: go to sleep after ai hours or after ai−1 hours." sounds inconsistent.

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for problem D, i Stored two arrays one array has (ai-bi) values and the other has (bi-ai) values and i ran a nested loop from i to n-1 & j=i+1 to n-1 and counted the number of elements that are greater than current element. what could i have done better?

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    3 weeks ago, # ^ |
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    In fact, you can first discrete all $$$a_i - b_i$$$ and $$$b_i - a_i$$$.

    Then use a Binary Indexed Tree to count all the values.

    The time complexity will be $$$O(n \log n)$$$.

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    3 weeks ago, # ^ |
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    Use a binary indexed tree.

    Insert $$$a_i - b_i$$$ into it after checking how many elements there are before $$$i$$$ which satisfies $$$a_j - b_j > b_i - a_i$$$. Discrete all $$$a_i - b_i$$$ and $$$b_i - a_i$$$ first thing first.

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3 weeks ago, # |
Rev. 2   Vote: I like it -31 Vote: I do not like it

UPD: Sorry for the comment. I didn't think much, it's my bad.

Appologize for the round, sorry.

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    3 weeks ago, # ^ |
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    Do you write is serious? The "subSTRING" problem, as you say, is not "more interesting" and can be solved in $$$O(n)$$$ considering $$$3$$$ or $$$4$$$ consecutive characters.

    D — data structure problem? The problem "sort the array and do (lower bound or binary search)" is data structure problem?

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      3 weeks ago, # ^ |
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      Sorry for that. I didn't think too much.

      I am terribly sorry for the words.

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      3 weeks ago, # ^ |
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      How to "sort the array and do (lower bound or binary search)"?

      It seems very interesting.

      I used binary indexed tree after discretion.

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    3 weeks ago, # ^ |
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    Can't the substring problem be solved in O(N)?, I did that only after I misread the problem! :( Just check all "3 consecutive characters substring" and "4 consecutive characters substring".

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Interesting fact for myself: Everytime people celebrate a rating-friendly contest, I lose ratings; Everytime people complain about toxic problems, I gain ratings.

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You did not think that the tasks are too simple even for div 3 ?)

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    3 weeks ago, # ^ |
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    Speed is really need. lol

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    3 weeks ago, # ^ |
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    vovuh made them easy to please div3 ppl

    here

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      3 weeks ago, # ^ |
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      Haha, that's funny. I don't know how to prepare "good" round. I prepare slightly hard round, people say "omg that's div.1, not div.3!". I prepare slightly easy round, people say "loool 2ez4me that was not even a round". I don't know what to do.

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        3 weeks ago, # ^ |
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        Lol. You should ignore comments made by candidate masters on div3 rounds. A candidate master saying div3 is easy is like a pupil / grey guy saying div1 is hard.

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        3 weeks ago, # ^ |
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        actually, vovuh, you did well, i really appreciate as most of div.3 contests are made of you

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        3 weeks ago, # ^ |
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        Don't coordinate anymore divs 3 rounds )

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        3 weeks ago, # ^ |
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        Actually, vovuh, you are the best. You prepare quality Div 3 contests, and I enjoy participating in your contests. Also, I participating out of the contest; I get to learn new concepts and techniques from your contests. I hope you keep up this great work and keep contributing to more such awesome contests.

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        3 weeks ago, # ^ |
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        I really liked D problem of this contest. I've been focusing on pbds/fenwick last 2 weeks and this problem really put my knowledge to test. I like how I saw 3 different approaches on this problem(bin search, pbds, fenwick and treap). The problem really managed to capture the creativity of the community. Well done, our Div.3 King

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        3 weeks ago, # ^ |
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        Great job vovuh, don't give in to the hate. As it is, it's hard and time consuming to create problems and test cases, let alone deal with criticism such as this.

        Personally, I think this is a good level of difficulty, otherwise how do we differentiate between Div 2 and Div 3 problems. Once in a while, problems in Div 3 are easier compared to other Div 3 contests — but I think many people in Div 2/Div 3 enjoy the opportunity to solve a few problems (that are in reach and where we can apply what we've learnt recently) and then learn from the other unsolved problems.

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    3 weeks ago, # ^ |
      Vote: I like it +9 Vote: I do not like it

    I think that this is the perfect difficulty level for Div 3. Myself being an Expert (Rating: 1734), I was only able to solve A, B, C, D. So, I believe that it justifies my claim.

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3 weeks ago, # |
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Speedforces

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3 weeks ago, # |
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Tried E for the first time in my life. Got wrong answer test 40! I think I made a silly mistake but couldn't figure out yet. My submission: 73079617

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    3 weeks ago, # ^ |
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    It failed for me as well, test case is involving l = 0 I guesss, and I had counted t = 0 case, which is wrong. Removing that gave me AC.

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3 weeks ago, # |
Rev. 2   Vote: I like it 0 Vote: I do not like it

Why these two submissions using PBDS for problem D got different verdicts Vovuh please look into the matter.

Correct SOlution:-73082017 Rejected Submission:-73079109

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    3 weeks ago, # ^ |
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    It's because, in your case, it couldn't contain same elements.

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3 weeks ago, # |
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During contest I was logged out several times. Does someone know why?

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3 weeks ago, # |
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It's quite easy. Even my i got more rating, but i think i wouldn't satisfy with those.

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3 weeks ago, # |
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What's wrong with this submission for D? https://codeforces.com/contest/1324/submission/73081429

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3 weeks ago, # |
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Why I am getting ILLEGAL CONTEST error while pushing "hack it"? Please fix

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    3 weeks ago, # ^ |
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    you must go to submission link and then press hack it there instead of hack from status

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3 weeks ago, # |
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Hey guys Can anyone tell me

Why this submission fails :https://codeforces.com/contest/1324/submission/73068802

and this one passes: https://codeforces.com/contest/1324/submission/73082736

I'm not able to find the problem can anyone point this out with an example.

Thank you!

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    3 weeks ago, # ^ |
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    Please guys :)

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    3 weeks ago, # ^ |
    Rev. 7   Vote: I like it 0 Vote: I do not like it

    ll cn1=pos.size()*(pos.size()-1)/2;

    pos.size() => returns (int value) not (long long value) you must work casting.

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      3 weeks ago, # ^ |
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      Thank you so much it worked now. Didn't work during contest so sad :(

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3 weeks ago, # |
Rev. 3   Vote: I like it +1 Vote: I do not like it

Hello? I'm a bit curious about how to solve F, can someone help me? Thanks

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3 weeks ago, # |
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can someone please explain the reason for TLE in my submission of E my code -link

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    3 weeks ago, # ^ |
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    Because at last where you have done dp[id][sum][f]=ans, your value of sum is changed, but your dp[id][sum][f] should use original value of sum not the updated one.

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3 weeks ago, # |
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1 5 2 2 2 2 4 why this will give yes in A?

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    3 weeks ago, # ^ |
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    NO, since all elements must have same parity (either odd or even), then only answer can be "YES"

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    3 weeks ago, # ^ |
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    Is it a full test or a single case? If it is a full test(i.e. 1 represents number of cases,5 represents number of columns) then simply put blocks in column 1-4 If it is a case(i.e. there are 7 columns),the answer is NO.

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3 weeks ago, # |
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+104 -> +8, thx for B pretests, goodbye expert:(

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    3 weeks ago, # ^ |
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    THIS HAPPENED WITH ME TOO :'(

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    3 weeks ago, # ^ |
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    exactly same problem.... :/ why would they create such weak tests?

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    3 weeks ago, # ^ |
    Rev. 2   Vote: I like it +1 Vote: I do not like it

    Actually, it means we should be more careful, and find out all special cases.

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      3 weeks ago, # ^ |
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      the thing is , the hacks arent even special cases, the hack could have easily been in the example test.

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        3 weeks ago, # ^ |
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        If you didn't consider something, for you it was a special case I suppose.

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          3 weeks ago, # ^ |
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          how did you figure out your rating change?

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            3 weeks ago, # ^ |
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            Chrome extension "CF-predictor"

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              3 weeks ago, # ^ |
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              Oh thanks. I can't see the prediction, I guess I was supposed to have it before the contest too?

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3 weeks ago, # |
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Hack for B?

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

!!NEED HELP!!

My solution for problem-D : 73076688

According to me,it's complexity nlogn, but i m getting time limit. Can anyone explain why it is happening or where m i wrong??

TIA

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    3 weeks ago, # ^ |
      Vote: I like it +1 Vote: I do not like it

    sum+=(ll)distance(it,ms.end()); has linear complexity -> O(n^2logn)

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      3 weeks ago, # ^ |
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      If I hack my own solution, then will that be an advantage?? like my scores would not reduce

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        3 weeks ago, # ^ |
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        No, you'll just hack your own solution instead of someone else.

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3 weeks ago, # |
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quite sad that my solution for B got accepted in the first place , was it intentional to make the test cases all weak for only an odd length palindrome / 2 different numbers only? would've solved 5 problems in one contest for the first time.

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    3 weeks ago, # ^ |
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    Yeah i mean i totally forgot about that test case.. I mean i knowingly did this when i should have not done it and it would have passed.

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3 weeks ago, # |
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Hi, can anyone explain why this solution 73084554 is giving WA. While this solution works 73037149 by tmwilliamlin168. Thanks in Advance!

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3 weeks ago, # |
Rev. 5   Vote: I like it 0 Vote: I do not like it

Problem B $$$O(N)$$$

73087070

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3 weeks ago, # |
  Vote: I like it +3 Vote: I do not like it

I liked this contest vovuh. Good job. I'm interested to know your thoughts on the difficulty levels of $$$D$$$ and $$$E$$$ if this was a Div $$$2$$$ contest. Do you think $$$D$$$ is harder than $$$E$$$ ?

How to solve $$$F$$$ ?

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    3 weeks ago, # ^ |
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    I think D is between the div2B and the div2C, while E is between the div2C and the div2D.

    However, you may just wait until when these problems are recorded in PROBLEMSET and the difficulty of them will be visble for you.

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      3 weeks ago, # ^ |
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      I think segment trees are too difficult to be asked for $$$B$$$. It might be a $$$C$$$.

      Can you please tell me how to solve $$$F$$$ ?

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        3 weeks ago, # ^ |
          Vote: I like it +8 Vote: I do not like it

        In fact, you don't need segment tree for D since the binary search is enough for it.

        For problem F, I apply DFS twice on the trees.

        In the first DFS, I regarded no.1 node as root and regarded the trees as a rooted trees. In this DFS, I get the correct answer for no.1 node.

        In the second DFS, I started my DFS from no.1 node and calculated the answer of other nodes in my traversal.

        For more details ( since I'm poor in English but good at C++ ), you can look up my code.

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          3 weeks ago, # ^ |
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          Nice ! I never thought of binary search for counting inversions ! :)

          My main doubt was in the second DFS. Thanks !

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          3 weeks ago, # ^ |
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          I implemented the same algorithm in Java, but failed test case 3. I couldn't figure out why it fails. Can you help me? My submission is: 73110580

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            3 weeks ago, # ^ |
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            It seems like ts3 is just a random tree whose node is all black. So you can just make a small test to see what happens in your program.

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              3 weeks ago, # ^ |
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              My tree building logic was wrong, I need to build a 0-rooted tree recursively, thanks anyway!

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    3 weeks ago, # ^ |
    Rev. 2   Vote: I like it +6 Vote: I do not like it

    Root the tree arbitrarily and let $$$dp_x$$$ be the answer for $$$x$$$'s subtree. We can do this with a single dfs.

    But then $$$dp_x$$$ won't necessarily have its full answer (because we didn't consider the path through $$$x$$$'s parent). So do a second dfs and for each parent node $$$y$$$ update each of $$$y$$$'s child's $$$dp$$$, (child may benefit from path via their parent $$$y$$$), but be careful to not to over count that child's contribution from $$$dp_y$$$.

    for clarity you can see my submission: 73067648

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      3 weeks ago, # ^ |
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      Ah ! The overcounting is tricky ! We should subtract only if $$$f(c) > 0$$$. The reason is that if $$$f(c) > 0$$$, then we never added $$$f(c)$$$ to $$$f(v)$$$

      Thanks !

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3 weeks ago, # |
Rev. 2   Vote: I like it +1 Vote: I do not like it

it is the first time i tried in E

but i got WA on 40

any one can help me why it is wrong

https://codeforces.com/contest/1324/submission/73078402

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3 weeks ago, # |
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3 weeks ago, # |
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Can someone please explain to me the DP state for problem E?

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    3 weeks ago, # ^ |
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    Before considering sleeps #$$$i,i+1,\ldots,n$$$, suppose you last slept at time t. Then under these conditions, dp[i][t] is the maximum number of good sleeps among $$$i,i+1,\ldots,n$$$. The question asks to find dp[1][0].

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3 weeks ago, # |
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Can B be solved in O($$$n^2$$$) will it pass?

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    3 weeks ago, # ^ |
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    i think. YES, as n=5000

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    3 weeks ago, # ^ |
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    Yes, but if you have a constant factor $$$a$$$ you will have runtime $$$O(a * n^2)$$$, which may TLE because worst case scenario you will do something like $$$a * 5000 * 5000$$$ operations which may not be executed in time (I hacked a couple ppl with this idea)

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    3 weeks ago, # ^ |
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    Yes, since sum of all n is <= 5000.

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Can someone explain why the order (i<j) won't matter in problem D ??

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    3 weeks ago, # ^ |
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    i<j simply means that pair (i,j) and (j,i) are same. so you can ignore this statement and divide it by 2 at last to consider it only once.

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    3 weeks ago, # ^ |
      Vote: I like it +2 Vote: I do not like it

    It is another way of telling that pairs (i,j) and (j,i) are the same. And they have to be counted as 1.

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3 weeks ago, # |
Rev. 2   Vote: I like it 0 Vote: I do not like it

Is problem F a classic problem? Because i notice that many people write similar code in their dfs. If it is, please tell me. Thanks is advance :)

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3 weeks ago, # |
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Could someone help me figure out my mistake in Problem E 73091842. Thanks in advance.

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3 weeks ago, # |
  Vote: I like it -10 Vote: I do not like it

Are you guys mad? I have missed the contest due to change in timing, now I can't attend div1 challenges.

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3 weeks ago, # |
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Vovuh you are great author your problems are amazing

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3 weeks ago, # |
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Hack case of B . 1 3 2 2 2

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3 weeks ago, # |
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how is this O(n^2) solution timing out on B for 2 secs??.Link

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    3 weeks ago, # ^ |
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    It seems like you are using map<int, int> which takes $$$O(\log n)$$$ time for adding and looking for element. Also, std::map is especially slow (big constant factor) so your solution has time complexity $$$O(n^2 \log n)$$$ with big constant factor.

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      3 weeks ago, # ^ |
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      Ah, wow. i always thought map was constant access and insertion. thanks btw

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        3 weeks ago, # ^ |
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        I think what you want is unordered map which uses hash. It has constant access time on average (but can be up to $$$O(n)$$$ worst case). It also is quite slow since internal structure is somewhat complicated. I mean, don't expect something like a array-access speed.

        Also, since it uses random, it can be hacked if some experienced coder tries to craft specific testcase to kill unordered map.

        I think there was a good blog on CF about this (and also how not to get hacked from that) but I can't find it now.

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      3 weeks ago, # ^ |
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      So this should time out too? 73028508

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        3 weeks ago, # ^ |
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        I don't think so. Your code is performing $$$n$$$ map access operations which make the whole complexity like $$$n \log n$$$. The code originally asked was using about $$$n^2$$$ map access.

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          3 weeks ago, # ^ |
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          Thanks. I was misinformed about the access time of maps, now its clear.

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    3 weeks ago, # ^ |
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    I think map takes O(log n) time for searching thus complexity goes to O(n^2logn)

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3 weeks ago, # |
  Vote: I like it +15 Vote: I do not like it

When it's my first time to solve 4 problems and I've a new high rank:

HACKS:

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3 weeks ago, # |
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Nice Contest! My solutions: A. Check the parity of all elements. If the parity is same, then the answer is YES, else NO B. Its always enough to check for palindrome sequences of length 3. So considering each element as center, check if the right and left parts contain atleast 1 element in common. C. its always beneficial to use R and not L. (why? bcoz with L you can only reach back to reach some R, which anyway you would have reached before reach the cell with L). so find the largest gap in reaching a R. D. Compute ci = bi-ai and sort this array. Then for each ai and bi count the number of elements in array c less than ai-bi. Also make sure to consider the edge cases where you have counted the same element again. E. Use dp. The state is (i, time), where i is the day (or the ith sleep) and time is the current time. Write transitions to each of ai and ai-1. F. Assuming you run bfs from some initial root node, Compute max(wi-bi) for each node i in its subtree rooted at i. Now we need to compute the value in parent part. This can be done using tree rerooting technique.

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3 weeks ago, # |
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Could someone help me with my problem E submission. I am not able to figure out the mistake.Thanks in advance[submission:73096035]

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    3 weeks ago, # ^ |
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    Corrected code 73127806

    Explanation
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3 weeks ago, # |
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Can anyone tell me why this 73097950 passes and this 73098017 does not ?

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    3 weeks ago, # ^ |
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    they will give different output when v contain same value more than once.Check this 3 6 6 6 2 2 2

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3 weeks ago, # |
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Can somebody hack this?

PS: I find similar solutions getting hacked.

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3 weeks ago, # |
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what's wrong? After submitting the (A) question of Round #627 Div. 3, in test case #2 it is showing this: "wrong answer Answer contains longer sequence [length = 100], but output contains 28 elements ". Can I know the reason?

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    3 weeks ago, # ^ |
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    You should print 100 output not 28, That is, 1 output for each test case.

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      3 weeks ago, # ^ |
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      I think my code is doing it. can you please check it? solution no: 73099642

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3 weeks ago, # |
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![ ](2-24510-trollface-deal-with-it-troll-face-png)

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3 weeks ago, # |
Rev. 2   Vote: I like it 0 Vote: I do not like it

I got TLE error during contest instead of WA! even after the contest when I was submitting correct answer I was getting TLE. I got frustrated and then copy pasted solution of another user then also I got TLE. Also, I was logged out multiple times automatically during the contest!! solution link: https://codeforces.com/contest/1324/submission/73104474

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3 weeks ago, # |
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Will there be any system testing after the hacking round is finished??

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    3 weeks ago, # ^ |
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    Yes

    All of the AC solutions will be rejudged with original tests and successful hacks once the hacking phase is over.

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3 weeks ago, # |
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Why I want to hack a solution and it shows Illegal contest ID ?

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    3 weeks ago, # ^ |
      Vote: I like it +6 Vote: I do not like it

    Try to hack solutions via the Status page instead of the Standing Page

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3 weeks ago, # |
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Can Somebody please explain why this solution is not giving compile time error on pretest cases and arbitrary cases on offline compiler Solution for Problem 2

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3 weeks ago, # |
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When system test will start?

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3 weeks ago, # |
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What's wrong with my solution for E problem? 73075794

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3 weeks ago, # |
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I am wonder that when the system test will begin.

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3 weeks ago, # |
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Anyone who solved problem D using two pointers? Please explain your approach

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3 weeks ago, # |
  Vote: I like it