By 300iq, 5 months ago, translation, In English,

Hi!

I'm glad to invite you to take part in Avito Code Challenge 2018 which starts on May/27/2018 17:50 (Moscow time). Any participant can join the round and it will be rated for each participant. Hope to see you among the participants!

Problems are prepared by me — Ildar Gainullin.

This round is conducted on the initiative and support of Avito. Avito.ru is a Russian classified advertisements website with sections devoted to general good for sale, jobs, real estate, personals, cars for sale, and services. Avito.ru is the most popular classifieds site in Russia and is the third biggest classifieds site in the world after Craigslist and the Chinese website 58.com.

Many thanks to Vladislav winger Isenbaev, Grigory gritukan Reznikov, Ivan isaf27 Safonov,Alexander AlexFetisov Fetisov and Shiqing cyand1317 Lyu for the round testing, Nikolay KAN Kalinin for helping me to prepare the contest, and also to Mike MikeMirzayanov Mirzayanov for systems Codeforces and Polygon.

Participants will be offered eight problems and three hours to solve them. Scoring will be announced a bit later.

Avito presents nice gifts for participants. Top 30 participants and also 10 random participants with places 31-130 will be awarded with a special T-shirt.

I hope everyone will find interesting problem for themselves. Wish everyone a successful round and high ratings!

Good luck!

UPD,Scoring: 500 750 1250 1750 2250 2750 3250 3750

UPD,Editorial is ready: Editorial!

The winners of random t-shirts are:

List place Contest Rank Name
6 981 36 snuke
31 981 61 Swistakk
40 981 70 yasugongshang
52 981 82 Xellos
62 981 92 shadowatyy
76 981 106 BudAlNik
80 981 110 kostka
88 981 118 pb0207
95 981 125 Noam527
96 981 126 xiaowuc1
Announcement of Avito Code Challenge 2018
 
 
 
 
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5 months ago, # |
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For a moment there I thought it was "Aviato", you know, Erlich's company.

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5 months ago, # |
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Will the difficulty of the problems be similar to other Codeforces Div1 & Div2 combined rounds?

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    5 months ago, # ^ |
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    Maybe the contest is including both Div.1 and Div.2 because it has 8 problems.

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is it rated?

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    5 months ago, # ^ |
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    This is a legitimate question because the blog says it's rated but the contest page doesn't say anything

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      5 months ago, # ^ |
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      It is given that it will be rated for every participant. See the first paragraph.

      And also the writer wished high ratings.

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        5 months ago, # ^ |
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        That was exactly my point the blog says it's rated but contest page doesn't say anything.

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    5 months ago, # ^ |
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    everybody here is downvoted?

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    5 months ago, # ^ |
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    Y!

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I want t-shirt. But I think i could not have this rank

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    5 months ago, # ^ |
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    I think you can get the T-shirt next time as long as you practice coding assiduously. Bless you.

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5 months ago, # |
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First Div 1+2+3 contest :D

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Cricket lover's in dilemma..#IPL final

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    5 months ago, # ^ |
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    May be miss the first innings of IPL final. But, it doesn't metter. Codeforces contest is more important than IPL final to me. :)

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5 months ago, # |
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Guess I won't be watching the IPL final then...

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when reading the Avito

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T-shirts will be awarded for 130 from Div.1 + Div.2?

Give Div.2 a chance :'(

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Will participants be able to do challenges(hacks)?

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    5 months ago, # ^ |
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    I think Yes, as he said that there is a score distribution for the problems (which will be announced a bit later). and he didn't say that it will be held on extented ACM ICPC rules.

    so this round will be like the ordinary combined rounds.

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Div1+Div2+Div3 contests at one time(8 problems) Yeah!!

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Karuis is looser, how it is possible!!!!!!!!!!!!!

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TIM_20180527081322

?

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    5 months ago, # ^ |
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    They do this for all not just for Mike Mirzayanov

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How many questions will there be?

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    5 months ago, # ^ |
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    Participants will be offered eight problems and three hours to solve them.

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T-shirts from Avito or from Codeforces?

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    5 months ago, # ^ |
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    Asking the real question. ^

    Not sure if I should try to reach for the T-shirt or wait for the next div2 round for a bigger rating increase kickstart into div1.

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The random t-shirts will be selected as usual with the help of two scripts below and random from testlib.h. Seed is the score of the winner in the upcoming Codeforces Round #485 (Div. 1), the length is the number of participants between the 31-th and the 130-th places inclusive.

python code
c++ code
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Delayed by 15 minutes :v

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what's the problem codeforces every round we get delay of 10-15 mins. at the last moment? :v

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Everybody now can watch IPL final for 10 minutes :)

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.

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    5 months ago, # ^ |
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    That second pic should say "after the contest" tbh fam.

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Whoever does not miss the Soviet Union has no heart. Whoever wants it back has no brain.

Vladimir Putin

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I do not participate in the contest but I discovered that the point decrement speed is not adjusted?

This way a problem will only worth 28% of original score at the end of contest.

Edit: just saw the announcement

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That's quite a lot of pretests. I like it (if they aren't mostly trash).

Not like I expect at least my F to pass, though.

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In D why is this approach wrong? (Wa on test 5)

I have dp[begin][end][cuts]

and if cut is 0 it is just the sum of begin to end

Else i can cut it at any of the location from begin to end and see the cuts from begin to cut_position and cut_position + 1 to end.

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    5 months ago, # ^ |
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    There is no guarantee that when choosing to cut at some point, the optimal AND value is obtained by the maximum possible values for the resulting two subarrays. Consider the pairs [5,5] and [5,2]. [5,2] gives us a better AND value (7), but your solution would choose [5,5].

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      5 months ago, # ^ |
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      How to solve D?

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        5 months ago, # ^ |
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        I did a bruteforcey solution with recursion every time I switch to a new bookshelf. With a few optimizations it passed in the time limit. (Mainly, don't recurse if the AND so far is less than the best cost you've seen so far) Update: Test case #61 begs to differ, got TLE

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        5 months ago, # ^ |
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        Start from the 55th bit, let now be 0. If you can get and value equal to now + 2^i, add 2^i to now. It can be done with easy dp.

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        5 months ago, # ^ |
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        Find the answer bit by bit. Start with answer equal to zero. Then for i from 60 to 0, check if you can obtain a solution with answer + 2i. If you can, just add.

        Checking if there is a solution with a fixed answer can by done with dp[position][parts]. To compute a state, try to fix a new position and consider dp[newposition][parts + 1] if the sum in the interval [position;newposition] contains the bits of the fixed value as a subset, in other words (sum&FixedValue) = FixedValue.

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          5 months ago, # ^ |
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          thanks for writing a neat and clear code. It help me a lot!

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        5 months ago, # ^ |
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        What I did:

        set<long long> dp[i][j] = The set of possible values obtainable when books are put into first i shelves, where first j books are used.

        It is trivial to show that in first i shelves, there should be i books in minimum and i+n-k books in maximum, and so the dp array is only valid in such range. A complete dp[i][j] can be computed by [all values in dp[i-1][i-1~j-1]] & [appropriate range sum of input array].

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    5 months ago, # ^ |
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    you got 1000 & 0100 < 0111 & 0001, when both 1000 > 0111 and 0100 > 0001

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I think G is too easy to be 3250.

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    5 months ago, # ^ |
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    I think the hardest part in solving G is debugging your code.

    PS: Btw, can anyone share with me how to debug segment tree code efficiently?

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      5 months ago, # ^ |
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      That's right, I forgot to mod the value after doing lazy updates. That costs me a lot of time and points.

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    5 months ago, # ^ |
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    how did you solve it?

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Can someone explain how to solve E?

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    5 months ago, # ^ |
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    Sort all queries and go from left to right. dp[i]- what is the last index where you can obtain i. When you encounter begin of any segment just update dp.

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      5 months ago, # ^ |
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      Sort the query by left, and if equal, by right?

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        5 months ago, # ^ |
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        By left. dp[x] = max(dp[x], min(dp[x-val_for_segment], end_of_seg))

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    5 months ago, # ^ |
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    I had a O((N / 64) * Q log(N)) solution with the offline dynamic connectivity strategy ( http://codeforces.com/blog/entry/15296?#comment-203606 ).

    Basically you do divide-and-conquer, keeping up a bitset representing the values attainable with currently active segments.

    At every step, you first make all segments that cover the current interval active, then recurse to the left half of the segment and to the right half of the segment.

    At bottom level, you OR the current bitset with the result bitset.

    You can add the points where a interval becomes active to a segment tree. Every interval becomes active at most log(N) times, so a interval becomes active Q log(N) times, so the complexity is O((N / 64) * Q log(N)).

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Can someone, please, explain how is updating the current set of active segments done in E? I couldn't reduce the complexity from O(Q2 * N / 64) for the whole second half of the contest. Or is the intended solution completely different?

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    5 months ago, # ^ |
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    I did it in , by dividing queries into blocks and processing offline. Its similar to handling delete queries in graph updating( maintaining connectivity, MST etc on edge updates) problems.

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      5 months ago, # ^ |
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      You can get the same complexity by performing a simple sqrt decomposition on the array and keeping bitsets with all possible values for each index.

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      5 months ago, # ^ |
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      I did it in by dividing queries better, like in the RMQ table or an offline segment tree, and going down from there. I tried and tried and couldn't see how to solve it more simply...

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        5 months ago, # ^ |
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        I have an O(NQ) solution but I'm not sure it's right or not.

        But if it is wrong, I would like to know a counter-example/why it doesn't work.

        Basically I do DP like how we do it without the ranges.

        Then I do sweepline, adding and removing values.

        When removing, I sort of undo by reversing the way how the dp is processed.

        http://codeforces.com/contest/981/submission/38675426

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          5 months ago, # ^ |
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          I basically did the same but with one large modulo instead of two small ones.

          Create events "segment starts" and "segment ends", order them by the coordinate, and maintain the knapsack array "f[x] is the number of ways to get x". Inverting the knapsack update loop is straightforward.

          If we stored the actual number of ways, this would certainly be a right solution. However, the actual number may be too large, so we store it modulo some number, which makes the solution fast but hackable with only moderate effort. Still, there were only few submissions for this problems per room, so the hacking did not occur.

          Edit: And this thread discusses the same solution.

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      5 months ago, # ^ |
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      Thanks! Time to actually learn sqrt-decomposition or some data structures, then.

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    5 months ago, # ^ |
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    I did it in o(n*n) with a 2d boolean array to check and a 1d integer array to count

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      5 months ago, # ^ |
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      How can you do it in O(1) if it takes O(Q) just to scan the input :/

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        5 months ago, # ^ |
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        Ops, i meant O(n * n), O(1) was for updating

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          5 months ago, # ^ |
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          How did you made the update in O(1)?

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            5 months ago, # ^ |
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            f[i][j] = true if you can get a sum equal i using value x[j], and d[i] is number of value x[j] take part in creating sum i. So if you remove a value x[j], set f[i][j] = false and decrease d[i] by 1.

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    5 months ago, # ^ |
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    My solution is completely different. For each query I stored the value xi in nodes in a Segment tree (each node stores a vector of values), and then took the bitset 10000000000... and pushed it downwards to the leaf nodes. The OR of all leaf nodes then tell us which values are possible and which are not.

    Complexity should be .

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      5 months ago, # ^ |
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      This is what I did in the actual contest (I didn't wanted to think more, had to code quick) but I think there are much nicer O(qn) solution.

      Suppose that all queries have Li = 1. You can do the following : we don't see DP[i] as a boolean array, but an integer array maximizing the minimum endpoint among all queries that it used. For each i, you can take j iff DP[j] ≥ i. This approach can be extended without restriction on L — do the "sweep" over increasing i, inserting query j when Lj = i. Similar problem is here.

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        5 months ago, # ^ |
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        I think I did exactly this approach. code

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      5 months ago, # ^ |
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      Looked into your solution, seems to be the same as mine, but with minor differences. Could you explain why is that we have such difference in perfomance? Mine got TLE..

      Here's my code:code

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        5 months ago, # ^ |
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        This part is wrong.

        for(int el:to_upd[v]){
            to_upd[v<<1].push_back(el);
            to_upd[v<<1|1].push_back(el);
        }
        

        Imagine that all Q queries are of the form (1, N, 1). Then you will push the 1s down, so that each to_upd[v] (with v being a leaf node) contain Q elements. And then the runtime will be O(N3) (N elements, each N queries with O(N) work for the bitwise OR).

        The idea to speed it up is the following: If to_upd[1] (root node) contains multiple elements, then you can immediately apply them to a bitset. This way you only have to do the operation bs |= bs << x only to_upd[1].size(), and not to_upd[1].size() * n. And similarly with the other nodes. Then propagate the bitset to the two child nodes, and not all values in to_upd.

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          5 months ago, # ^ |
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          Thanks, I forgot that the updates are duplicating.

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      5 months ago, # ^ |
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      can you please explain your solution a little bit more.

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        5 months ago, # ^ |
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        Check out my solution: 38674330 I think my code should be easy to understand.

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          5 months ago, # ^ |
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          thanks for sharing this :) Btw if possible then can you please explain the second idea also!

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    5 months ago, # ^ |
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    I have :

    Keep dp[x] — count of subsets with sum X mod 109 + 7. Then we can easily add and remove numbers in .

    Here is the code.

    PS: It passed so there were no anti-MOD tests.

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      5 months ago, # ^ |
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      Why can you delete like that? Couldn't something be added (or deleted) in between that influences dp[i], but not dp[i + x]?

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        5 months ago, # ^ |
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        I also had some trouble figuring this out. But now I got it.

        You can think of it in this way: dp[i+x] consists of the ways that include the summand x, and of the ways that don't include the summand x. Let's say dp[i+x] = dp_with[i+x] + dp_without[i+x]. The goal is to compute dp_without[i+x], which is dp[i+x] - dp_with[i+x].

        Now, dp_with[i+x] is clearly the same as dp_without[i], since we can make a bijection between each combination with the sum i without the summand x and each combination with the sum i+x with the summand x.

        Now, notice that dp_without[i] = dp[i] for all i < x, and because radolav11 computes the values with increasing i, he can just ignore the helper array dp_without and store the result directly in dp.

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      5 months ago, # ^ |
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      There were anti-mod tests, though not for 1e9+7. I just let the dp[x] value overflow since it anyways takes mod with 2^31, but it failed on test 75

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      5 months ago, # ^ |
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      I passed it with 1e9+9. Is there a way to do something similar with no mods?

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How to solve G?

I tried to do it with O(n log(n)^2) segment tree beats but got TLE'd (pretest 15). Is there a better way?

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    5 months ago, # ^ |
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    I don't remember what segment tree beats are (other than that the name evokes someone beating a segment tree with a stick), but my solution was keeping sets for all intervals containing x for each x, using it to extract intervals of multisets to multiply by 2 and intervals of multisets to add 1 to; these 2 types of queries (+ get sum) can be handled with a fairly normal lazy segment tree. since there are only O(Q) intervals to update.

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      5 months ago, # ^ |
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      Okay, thanks.

      I did the same but stored the intervals that have all nodes containing x in the segment tree, not in a separate set -> * log(n) more intervals -> TLE.

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      5 months ago, # ^ |
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      Xellos, Could you explain how to deal with the two operations in the lazy segment tree? I don't know how to keep it log(n) per query.

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        5 months ago, # ^ |
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        Keep sum, mul and add for each segment.

        • when multiplying a segment by x, multiply sum, mul and add by x
        • when adding x to a segment, add x to add and size·x to sum, where size is the number of multisets in the segment
        • when sending mul and add down to sons, first multiply them by mul and then add add to them
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          5 months ago, # ^ |
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          Thank you! That makes a lot of sense.

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How to Solve 3rd Question , Tree Decomposition .

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    5 months ago, # ^ |
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    Only the stars are valid for that problem

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    5 months ago, # ^ |
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    If it is linked list — the answer is the list itself.
    Otherwise there must be only one vertex with degree > 2. This vertex joins all other paths.

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      5 months ago, # ^ |
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      Linked list is a pretty extraordinary expression for path :p (some people call it bamboo, though xD)

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      5 months ago, # ^ |
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      check the degree of all vertex , if there are more than 1 vertex with degree 2 than No.

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        5 months ago, # ^ |
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        You mean more than 1 vertex with degree greater than 2

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jtnydv25, name is enough to understand who is he?

one of my most favourite coder...

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Why do tasks on Codeforces have to be like this http://codeforces.com/contest/981/problem/G and this http://codeforces.com/contest/981/problem/H instead of like this https://agc024.contest.atcoder.jp/tasks/agc024_e and this https://agc024.contest.atcoder.jp/tasks/agc024_f >_>? Kinda obvious what to do more or less and then codecodecodecodecodecodecodecodecodecodecodecodecodecodecode and carefully work out details

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    5 months ago, # ^ |
    Rev. 2   Vote: I like it +39 Vote: I do not like it

    Sometimes I also wonder how problem writers on ATCoder can come up with such creative and original tasks (which is a rare thing on most of other online judge). Can any writers there share your experience?

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    5 months ago, # ^ |
      Vote: I like it +32 Vote: I do not like it

    It's "Code Challenge", you know.

    I agree problems were on the easier side, but I don't really agree that they were hard to code.

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5 months ago, # |
  Vote: I like it +6 Vote: I do not like it

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5 months ago, # |
  Vote: I like it 0 Vote: I do not like it

Love these statements

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5 months ago, # |
  Vote: I like it +5 Vote: I do not like it

Simple O(N2) solution to A is pretty cute. I don't know why author gave N <= 50 and ruined the fun :p

And I really liked problem F. thanks to authors!

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    5 months ago, # ^ |
      Vote: I like it +54 Vote: I do not like it

    Isn't the answer either 0, n or n-1? That would lead to trivial O(n). Or am I mistaken?

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      5 months ago, # ^ |
      Rev. 2   Vote: I like it 0 Vote: I do not like it

      Now I see that too. Too cute to be true :)

      I wrote it after finding a solution in my room, that only check prefixes.

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5 months ago, # |
  Vote: I like it 0 Vote: I do not like it

Is it really necessary to have 130+ tests for A?

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    5 months ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    Maybe most of them are from successfull hacks.

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    5 months ago, # ^ |
    Rev. 2   Vote: I like it -8 Vote: I do not like it

    If there is at least one solution that fails from it, then yes they are neccesary =)

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5 months ago, # |
  Vote: I like it +110 Vote: I do not like it

What kind of pretests are these XD?

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    5 months ago, # ^ |
      Vote: I like it +30 Vote: I do not like it

    LOL

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    5 months ago, # ^ |
    Rev. 2   Vote: I like it +14 Vote: I do not like it

    I hacked 2 such solutions in my room. I was shocked when I saw these and rechecked the question to see if I had read it right.

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    5 months ago, # ^ |
      Vote: I like it +24 Vote: I do not like it

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5 months ago, # |
  Vote: I like it +41 Vote: I do not like it

How to solve F? First binary search, then reduced it to some matching problem with circular segment, then what?

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    5 months ago, # ^ |
      Vote: I like it +81 Vote: I do not like it

    I solved it by using Hall's theorem. Check segment instead of subset is enough!

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    5 months ago, # ^ |
      Vote: I like it +41 Vote: I do not like it

    Here's a more detailed editorial: binary search on the answer. For each bi, expand it to an interval ci = [bi - x, bi + x]. Now we need to see if there's a matching between the ai and these intervals ci where a point and an interval can be matched if the interval contains that point.

    If this were a line, it would be easy: sweep through the intervals left to right, pick the first point in the interval. However, we're on a circle, so intervals that cross the origin are tricky. To simplify it, "unroll" the line and double it. If an interval was originally [l, r] with l ≤ r, then add intervals [l, r] and [l + L, r + L]. Otherwise, just add the single interval. Now we have at most 2n points and intervals and run the linear matching algorithm as described above. Hall's theorem tells us that this unrolling is equivalent to a matching on the circle.

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      5 months ago, # ^ |
        Vote: I like it +61 Vote: I do not like it

      Thanks for the details! I don't really get why Hall's theorem says that the unrolling is equivalent. Could you elaborate?

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        5 months ago, # ^ |
          Vote: I like it +34 Vote: I do not like it

        Sure.

        Initially, we have two sets of length n, A = {a1, a2, ..., an} and B = {b1, b2, ... bn}. Now when we do this unrolling process, we get two new sets A' = {a'1, a'2, ... a'n, a'n + 1, ... a2n} and C = {c1, c2, ..., cm} where a'i = ai%n and ci = [bi - x, bi + x]. Note that we use m instead of n because bi might map to two ci. n ≤ m ≤ 2n.

        We have a perfect matching between A' and C. Let's apply Hall's theorem. Consider some set S', which is a subset of these a'i. Then because we have a matching, |S'| ≤ |N(S')|, where N(S') is the set of ci that are connected to some node in S'.

        Now, to prove that there's a perfect matching between A and B, we use Hall's theorem in the other direction. Choose some subset of nodes in A, call it S. Then what is |N(S)|? Let's define S' as a subset of A' having BOTH copies of ai for every ai in A. Then we know by the previous paragraph that |S'| ≤ |N(S')|. Since 2|S| = |S'| and 2|N(S)| ≥ |N(S')|:

        |S| ≤ |N(S)|, so there's a matching in the original graph.

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          5 months ago, # ^ |
            Vote: I like it +22 Vote: I do not like it

          I think it's wrong proof.

          1) Forget about the case, when |C| = 2 × n, because in this case there are no intervals, that crossing the origin and this case can be easily solved, as mention above.

          2) But if you have at least one crossing the origin interval, that it means, that |C| = m < 2 × n. After that you are not able to have perfect matching, because the sizes of components are different.

          3) So, I suppose you want to say, that you have the matching, where one part is covered by matching full. But again you told: "Consider some set S', which is a subset of these a'i. Then because we have a matching, |S'| ≤ |N(S')|, where N(S') is the set of ci that are connected to some node in S'.". If you take S' = A', than you have: , but , contradiction.

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

For ques B

Is this a valid test case ?

4
1 10 
2 6
3 5
4 2
4
1 1
2 2
3 7
4 8

If yes

what is the answer for this test case

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    5 months ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    The answer should be 31 = 10 + 6 + 7 + 8 = max(10, 1) + max(6, 2) + max(5, 7) + max(2, 8).

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      5 months ago, # ^ |
        Vote: I like it 0 Vote: I do not like it

      But in the statement its given that

      "There shouldn't be equal elements in the subsets."

      In this case both the subsets have 2 elements.

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        5 months ago, # ^ |
          Vote: I like it +1 Vote: I do not like it

        That's "equal elements", not "equal number of elements" .

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          5 months ago, # ^ |
            Vote: I like it 0 Vote: I do not like it

          Oh, i thought that was stated in first statement of the problem.

          I took it for equal number of elements.

          Thanks for clarifying.

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        5 months ago, # ^ |
          Vote: I like it +1 Vote: I do not like it

        Well, this means that we should choose some elements from array a[], some elements from array b[], and no one from chosen elements are the same.

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5 months ago, # |
  Vote: I like it +81 Vote: I do not like it

modulo 998244353

I see this line and I feel sick. I see this line in two problems and I want to leave competitive programming for another year.

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    5 months ago, # ^ |
    Rev. 3   Vote: I like it +38 Vote: I do not like it

    I guess H is FFT problem so author wants to hidden it by letting that modulo in G :)

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    5 months ago, # ^ |
      Vote: I like it +97 Vote: I do not like it

    On contrary, I always pay respect to problemsetters who put this modulo in their problems. By putting it in problems where any modulo is sufficient we obfuscate which problems need FFT.

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      5 months ago, # ^ |
        Vote: I like it -35 Vote: I do not like it

      I think giving 109 + 7 in every single problem, including FFT ones is better way to obfuscate it. No one should use NTT anyway.

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        5 months ago, # ^ |
        Rev. 2   Vote: I like it +24 Vote: I do not like it

        Giving 109 + 7 makes it harder to give larger constraints like 105

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          5 months ago, # ^ |
            Vote: I like it -6 Vote: I do not like it

          Then the model solution just sucks, author should come up with better algorithm.

          I don’t understand how can mod 998244353 ever be a good idea. The author requires large constant factor to solve the original problem, and to overcome it they use some peculiar modulos that is known to be useful for a limited number of audiences. For someone to easily solve that problem, they should knew that “magic number” beforehand, and the fact that they have good primitive roots. Sounds pretty bad to me.

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            5 months ago, # ^ |
              Vote: I like it +25 Vote: I do not like it

            This situation is not much different than simply knowing NTT at all.

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

          I doubt that. Splitting numbers in two groups is still faster than making computations modulo some number. And it works for every mod.

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            5 months ago, # ^ |
              Vote: I like it 0 Vote: I do not like it

            Karatsuba works for every mod, doesn't need too many modulo operations, has a nice constant, works exactly (no doubles) and if the use of FFT is complex enough — the problem isn't just reduced to a single FFT or one of garden variety D&C+FFT versions — the constraints are often long enough that Karatsuba can pass. Its only disadvantage is asymptotic complexity. Actually, even D&C+Karatsuba has complexity equal to Karatsuba only.

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              5 months ago, # ^ |
                Vote: I like it +6 Vote: I do not like it

              Well, Karatsuba is nice algorithm indeed :)

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            5 months ago, # ^ |
              Vote: I like it +31 Vote: I do not like it

            I disagree. Modulo is faster than splitting and then doing computations on pairs of doubles.

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        5 months ago, # ^ |
        Rev. 2   Vote: I like it +49 Vote: I do not like it

        I don't like problems that require FFT with 109 + 7 modulo. Why is it better? Only because simple fractions like looks nicer? Not a big argument.

        NTT has much shorter code, which is important in ACM-style contests, where you don't have any prewritten code, and it works absolutely with same speed. When I tested FFT vs NTT, difference in execution time was about 20%, and one of them was faster on G++, and another was faster on mingw, so they are absolutely equal in my opinion (only FFT requires slightly more memory, which can be critical in some cases).

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          5 months ago, # ^ |
          Rev. 2   Vote: I like it +4 Vote: I do not like it

          I think I'll continue giving 109 + 7 in my FFT problems though :)

          It is better because it forces you write more generic solution and doesn't indicate key idea just from mod (giving NTT mod in every problem is even worse, in my opinion). You can use NTT with two primes + Chinese remainder theorem if you dislike that. It's much slower though...

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            5 months ago, # ^ |
              Vote: I like it +129 Vote: I do not like it

            Good thing is that we usually know the authors before the contest

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              5 months ago, # ^ |
                Vote: I like it +21 Vote: I do not like it

              Yes-yes, you skip every my contest, I know.

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5 months ago, # |
  Vote: I like it +9 Vote: I do not like it

I saw a possibly incorrect submission (see 38668742) for 981D - Bookshelves, while I have no idea how to hack it during the contest time.

The solution is simple, for dp[index][count], maintain the top 20 bitwise-and sum that ends at index with count blocks. It seems true to me that there exist cases which fail this solution(cases which you need some smaller and-sum in the middle-way).

So, can anyone suggest a way to generate the hack? Or there are some ways to prove that this solution is indeed correct?

PS. I hope these kind of submissions will fail systest... I mean, I hope the systests are strong :D

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5 months ago, # |
  Vote: I like it +8 Vote: I do not like it

Can someone explane me how I got WA on test 5 in final testing and someone got WA on test 13 during the contest ?

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    5 months ago, # ^ |
      Vote: I like it +19 Vote: I do not like it

    As far as I know the first couple of actual tests aren't always the same as pretests. Your mistake is "dfs(1,0); -> dfs(last, 0)". I know about some people who hacked with a similar test so the test wasn't in pretests.

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      5 months ago, # ^ |
        Vote: I like it 0 Vote: I do not like it

      Thanks !

      I saw mistake, but situation with testcases was strange to me :)

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5 months ago, # |
  Vote: I like it 0 Vote: I do not like it

What can be the cause of Wrong Answer for this solution of D ? Link:

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    5 months ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    Same Problem Also wanna know why my code is wrong. My Solution of D

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      5 months ago, # ^ |
        Vote: I like it +6 Vote: I do not like it

      you take the max at each step, but it is not the best choose at this step

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        5 months ago, # ^ |
          Vote: I like it -8 Vote: I do not like it

        why how ?? what's the alternative one. and why this approach is wrong ?

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          5 months ago, # ^ |
            Vote: I like it 0 Vote: I do not like it

          Max(7, 8)&3=0 Min(7, 8)&3=3 in such solutions you need to remember all the answers and update the new answer with them

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            5 months ago, # ^ |
              Vote: I like it 0 Vote: I do not like it

            ohk got it. That's very silly one. thanks

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

I have no idea about the complexity of my solution for G. Only that is not bigger than O(n * q)

Link

But I got AC))

I have one main segment tree for calculating the answer and also store additional segment trees for each of the values 'x'.

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    5 months ago, # ^ |
      Vote: I like it +8 Vote: I do not like it

    It's not necessary that a range modification had common parts with some modified ranges before, because after the current modification, these ranges would merge together, so the total time complexity is

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      5 months ago, # ^ |
        Vote: I like it 0 Vote: I do not like it

      I thought about this, but they will be merged only in this 'x'.

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        5 months ago, # ^ |
          Vote: I like it +5 Vote: I do not like it

        The different 'x' doesn't affect the time complexity. One range modification will only affect at most one range after.

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5 months ago, # |
  Vote: I like it +11 Vote: I do not like it

Hi everyone.

For problem D , Initially I read like, we can divide 'K' subsets in any order. (Not necessarily contiguous segments of array) .

Is it really possible to solve it ?

In last 25 minutes I read it again, and figured out Greedy-DP solution. which seems like HACK to me. And still the solution passed. I was hoping this solution to fail. Can someone tell me, is it possible to hack this solution ?

Basically I am picking up best 8000 values for each dp[i][j] , where 'i' is index, and 'j' is number of shelf's being used till index 'i'.

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5 months ago, # |
  Vote: I like it +29 Vote: I do not like it

I enjoyed the problems, even if I didn't well. Thanks to the author.

Out of curiosity, am I the only one who found problem C easier to understand by translating the Russian version to English with Google Translate?

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5 months ago, # |
  Vote: I like it +24 Vote: I do not like it

How to solve problem F?

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5 months ago, # |
  Vote: I like it +207 Vote: I do not like it

No test cases with 1<k<n-1 for H? Top two both fail with such k...

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    5 months ago, # ^ |
      Vote: I like it +182 Vote: I do not like it

    Wow, indeed :) My solution fails due to an incorrect assert(cur == 0);, removing it fixes the trouble. Perhaps I should be more careful with my assertion checks next time...

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    5 months ago, # ^ |
      Vote: I like it +9 Vote: I do not like it

    WAAAAAAT XDDDD? Oh my holy Mike...

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    5 months ago, # ^ |
    Rev. 2   Vote: I like it -15 Vote: I do not like it

    Lol, nice catch! Also: xD

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    5 months ago, # ^ |
      Vote: I like it +41 Vote: I do not like it

    Oops! Sorry, i thought these testcases are useless :/

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    5 months ago, # ^ |
      Vote: I like it +35 Vote: I do not like it

    Do you know what hour is it? It's rejudge hour!

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      5 months ago, # ^ |
        Vote: I like it +6 Vote: I do not like it

      People might feel like u r crazy , because, u want re-judge.

      but they don't know, u would have rank-1 , if there would have been re-judge :P .

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5 months ago, # |
Rev. 2   Vote: I like it -77 Vote: I do not like it

can some one suggest whats wrong in this code for D, thanks in advance ~~~~~

include<bits/stdc++.h>

using namespace std;

typedef long long ll;

ll dp[55][55];

ll ar[55]; const ll inf=(1LL<<60)-1;

int n,m;

ll foo(int i,int j)

{

if(j>m)

return 0;

if(i==n)
{
   if(j==m) 

   return inf;

   else

    return 0;
}
if(dp[i][j]!=-1)

return dp[i][j];

ll p=0;

ll ret=0;

for(int k=i;k<n;k++)
{
    p+=ar[k];

    ret=max(ret,(p&foo(k+1,j+1)));
}
return dp[i][j]=ret;

}

int main() {

memset(dp,-1,sizeof(dp));
cin>>n>>m;
for(int i=0;i<n;i++)
{
    cin>>ar[i];
}
cout<<foo(0,0);

}

~~~~~

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    5 months ago, # ^ |
      Vote: I like it +23 Vote: I do not like it

    When you have a problem that involves AND operation is a bad idea to use max because not necessarily if you return a max number you guarantee a better answer, for example:

    • 11 & 4 = 0
    • 11 & 3 = 3

    What happened here? 4 is greater than 3 but if we return 4 we get the worst answer.

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5 months ago, # |
  Vote: I like it +39 Vote: I do not like it

Your crafting.oj.uz ratings are updated!

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5 months ago, # |
  Vote: I like it +7 Vote: I do not like it

Will there be editorial for this contest?

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5 months ago, # |
  Vote: I like it 0 Vote: I do not like it

Can anybody please explain their approach for problem D.

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    5 months ago, # ^ |
      Vote: I like it +8 Vote: I do not like it

    A observation first: We will see if we can obtain in the final sum, every bit from 60 to 0. (We must find some configuration such that every shelf contains in the sum of the values that bit)
    If we can obtain a bit, then we add it to the answer, and when we try for a new bit, we always respect the bits that are already found.

    I have solved it using dynamic programming dp[i][j][h] -> true or false, if we can put the first i books into j shelves such that each shelve has the bit h in its sum. (And also respects all the higher bits found previously)

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      5 months ago, # ^ |
        Vote: I like it 0 Vote: I do not like it

      I am not clear with this part "And also respects all the higher bits found previously" Can you please explain what do you mean by this line

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        5 months ago, # ^ |
          Vote: I like it +14 Vote: I do not like it

        Let's say for example that you are at bit 5, and you know that you have found good configurations that can have bits 8, 10 and 15. (The answer until now is 28 + 210 + 215). When computing dp[i][j][5] we have to find some index i2 (smaller than i) such that sum(i2, i) contains 25 + 28 + 210 + 215, and dp[i2][j - 1][5] is also true (which means that the first i2 books can be put in j-1 shelves such that their total sum contains 25 + 28 + 210 + 215)

        My code: 38668315

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5 months ago, # |
  Vote: I like it +35 Vote: I do not like it

Where is editorial's link ??

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5 months ago, # |
  Vote: I like it 0 Vote: I do not like it

Thanks for great contest!

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5 months ago, # |
  Vote: I like it 0 Vote: I do not like it

How to solve H? Where's the solution?

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5 months ago, # |
  Vote: I like it +1 Vote: I do not like it

How to solve D ? Cannot understand the approach listed in the comments

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5 months ago, # |
  Vote: I like it 0 Vote: I do not like it

Editorials please??

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5 months ago, # |
  Vote: I like it -13 Vote: I do not like it

How about the Editorial????

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5 months ago, # |
  Vote: I like it -15 Vote: I do not like it

Editorial ??

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5 months ago, # |
Rev. 2   Vote: I like it +35 Vote: I do not like it

Hello

English editorial will be ready in 2-3 days, sorry for inconvenience, i am busy with some school exams now

Now you can translate russian version with some translator like yandex

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5 months ago, # |
  Vote: I like it -19 Vote: I do not like it

Where are the editorials?

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5 months ago, # |
  Vote: I like it -14 Vote: I do not like it

When will the tutorial be uploaded?

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5 months ago, # |
  Vote: I like it -14 Vote: I do not like it

When will the editorials be posted?

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5 months ago, # |
  Vote: I like it -13 Vote: I do not like it

can someone direct me to the link of the contest editorial for this ?

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5 months ago, # |
  Vote: I like it 0 Vote: I do not like it

Does anyone have an editorial for the problems in this contest :(

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5 months ago, # |
Rev. 2   Vote: I like it 0 Vote: I do not like it

Can anyone please help me find a mistake in my solution for D.

Here is what I did:

dp[s][e][j] — stores the maximum possible beauty of j shelves when they contain books from as to ae

if(j == 1)
     dp[s][e][j] = sum(a[i]) for i = s to e
else if(e-s+1 == j) means j books in j shelves
     dp[s][e][j] = bitwise and (a[i]) for i = s to e
else 
     dp[s][e][j] = max(sum(a[s]...a[i-1]) & dp[i][e][j-1]) for i = s+1 to e-j+2

The answer to the problem is dp[1][n][k]

Here is my code.

Thanks in Advance

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    5 months ago, # ^ |
    Rev. 2   Vote: I like it +16 Vote: I do not like it

    First,s = (s&a[i]);must be s = (s&a[j]);.

    Second,This test case:

    4 3 11 6 1 3

    Answer:3.Your output:0

    11&(6+1)&3=3.But in your solution,dp[2][4][2]=4.(6&(1+3)).

    It means that it not necessary to use the maximum value in[L,R].

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      5 months ago, # ^ |
      Rev. 2   Vote: I like it 0 Vote: I do not like it

      ninja'd

      Btw, there is also a second error in the andd computation. The for loop needs to start j = i.

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    5 months ago, # ^ |
      Vote: I like it +6 Vote: I do not like it

    This problem is not greedy in that way you assume. dp[s][e][j] = max(sum & dp[i][e][j-1]) doesn't always hold.

    E.g. take a look at the following example. If you want to make two bookshelves using the books [7, 1, 12], then the best you can do is (7+1) & 12 = 8. However if you add a book [4, 7, 1, 12] and want to make three bookshelves, then the best you can do is 4 & 7 & (1 + 13) = 4. Here you have to take a smaller value for the subset [7, 1, 12], namely 7 & (1 + 12) = 5, to get a bigger end result. Because otherwise 4 & (7 + 1) & 12 = 0.

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

In China. The contest in provinces is called NOIp. And we have two tests. The first test is printed on papers and one part is to read a program and write what it outputs. And then when I was taking a mock test and doing the part. I see the following four lines of the code:


/** * author: tourist * created: 27.05.2018 18:26:43 **/