Hello everybody!

The Technocup 2017 (a competition for high-school students hosted by Mail.Ru Group together with MIPT, Bauman MSTU and Codeforces) Finals is happening tomorrow in Moscow. For those who will not participate in the Finals, we prepared mirror contests for both divisions. The round will start at 13:05 UTC at Sunday, March 5th, 2017 and will last for two hours.

The round was prepared by Endagorion, WHITE2302, Alladdin, fcspartakm, Amethyst1, MikeMirzayanov, ifsmirnov and me.

You can view the current onsite results at link.

There will be six problems in each division.

Editorial is here.

Why isn't this on the home page?

Will there be an editorial for Div2 and Div1 problems ?

What makes you think I would know this? xP

Sorry, I just hijacked the top comment xD

http://codeforces.com/blog/entry/50854

I thought that I will die before I participate in a new round

hahaha!

Error 403: this page is not available in your country...Syrian contestants will relate :( :( :(

I have developed a habit of viewing if there is a new round every day...

*every hour

Will the descriptions of the problems in English or in Russian? Thanks a lot.

both in Russian and in English

wish in chinese

so am i.

so am I.

Scoring distribution ?

Will it be rated or unrated?

Yes, it will. In Russian it is stated clearly.

@Andrekk, Could you please give me the link of the russian statement?

I can't do it. You switch the language via button in the upper right corner of any web page, but URL isn't changed. Basically, in russian it says, ' А для тех, кто не будет участвовать в финале, мы подготовили рейтинговую трансляцию финала для обоих дивизионов' which was translated just like 'For those who will not participate in the Finals, we prepared mirror contests for both divisions'.

Slept for 2 hours, went for an ACM training, then a CF round... feel awesome!

What are the problem scores?

It'll be announced after the contest begins. :D

scores distribution?

Rating prediction for this contest could be found for div1 & div2

Since today results table sortable, just click on column header to sort by it.

Extensions:

About CF-Predictor

Good luck & high rating for everyone!

Amazing! Thank you.

What is

seed? My position in sorted list of competitors ordered by their rating?Seed is your expected ranking according to your current rating. In other words, if your final ranking equals your seed, then your rating change should be 0. If your final ranking is lower than your seed, then you will gain rating points.

Thank you. How is that determined?

http://codeforces.com/blog/entry/102

http://codeforces.com/blog/entry/20762

Thank you.

You are welcome:) If talk simple, if your rank equals to your seed, your rating doesn't change. But actually it could change according to anti inflation policy.

Very weak pretests in problem D. LOL :D

OMG!! MY SOLUTION IS CORRECT!! HOW?! EVEN I DIDN'T READ STATEMENTS FULLY!

Your submission : http://codeforces.com/contest/782/submission/25256168, Should have Failed on : Weak systest

My answer is NO for this case.

Yeah, it should be :

Is my solution correct? 25263983

I am getting WA on pretest 5, for Div2 C. Can anyone help me? My submission: 25264506

Please remove your code, it is covering page, just keep submission link please.

Done.

I thought that block of code means one can see code only if he clicks on the top of codes, but I was wrong. So I have removed my code.

25270511 Just a small change in dfs checking if val(parent)==id is required after checking for the current node as well and AC! :D

Thank you pratik.pugalia I have now found my mistake.

Thanks a lot for your comment !

Can you solve Div2B using binary search? I tried searching for the time or for the meeting point, but I'm getting WA6. Is the approach wrong or I just screwed up floating-point arithmetic?

I had the same problem. After 5 unsuccessful submissions, I put cout.precision(7); at the beginning of the code and it passed the pretests.

Damn, that did it, I got AC. I feel really stupid now, I'm not even sure why this line matters.

Thanks.

It's because by default cout precision is low. The problem requires precision upto 10^-6. So, suppose answer is 1.000777, cout without precision setting would round it to 1.001, resulting in WA.

Don't we need ternary search?

No. A unimodal function can't be solved with binary search. You need ternary search.

You can binary search on time, it's monotonic.

yeah, it's monotonic on time, unimodal on distance.

How to prove its unimodal on distance ?

I used Ternary search and got AC. It should not be Binary search bcz then we may choose boundary values(min or max).

Edit:it's possible follow this (binary search on time)I used ternary search and got TLE.

Ternary search works perfectly fine. Got AC. Check for boundary conditions. Using iterative method works. Compute log2(MAX_NO_OF_ITERATIONS) and do the search that many times.

CodeFunction fPlease look into my submission.

http://codeforces.com/contest/782/submission/25253601

i think it's due to eps use 1e-6 http://codeforces.com/contest/782/submission/25268096

You may check my solution.Binary search works just fine.I haven't checked a ternary search solution yet,but I believe these two methods are both correct,though may be different in some details.Code

Test case 6 in Div 2 B ?

How did you solve Div2 B?

Binary search the answer.

Binary search a suitable x and find t, take the minimum of all iterations.

Binarysearch

Is it possible by applying binary search on distance?

i didn't get, can you explain a bit more about canmeet function ,please?

I'll try to explain.

In some time 't', any person can be present at positions in range (x[i]-t*v[i] to x[i]+t*v[i]).

All the persons will be present at a common position 'p' iff 'p' lies in all such segments.

How to solve B? I think this it can be solved by ternary search, but I couldn't implement.

I passed pretests using ternary search. I'm not sure it is a correct solution, though.

I make it 10.

Suppose now we are searching (l,r). Denote t=(r-l)/10, we divide (l,r) into 10 smaller sections(l, l+t), (l+t, l+t*2) ......(l+t*9, r), then compute the result of l+t, l+t*2, l+t*3 .....l+t*9. Suppose l+t*k is the minimum, the optimal answer must be in (l+t*(k-1), l+t*(k+1)).(10 sections -->2 sections）

After trying to find a flow solution for problem B, my head is now overflowed...

PS: How to build the flow correctly for problem B?

B -- This can be modelled as a maximum bipartite matching, with all teams sharing the first option having all their first options removed. This is a standard algorithm -- See Hopcroft-Karp for an efficient algorithm.

I feel like an idiot after reading your solution. That's a lot simpler than I thought.

It can be modeled as 2SAT also, but both of these are overkill solutions. I saw some people in my room had done some nifty 30 line solutions...

I bet that all these nifty solutions look like flow algorithms. Nson's for example:

All teams start with the first option (xxx). Until there's no conflict for some team

i, make teami's choice the second option (xx+y).That's it. This is the solution. And it reminds me very much the idea "find an augmenting path and send flow through it" (Ford-Fulkerson's).

Can you please explain your graph weights? I tried thinking on the flows solution during the contest, but was unable to find an analogy which completely fits the solution.

This is a maximum bipartite matching (not exactly max flow), but the most efficient popular solution (in contest contexts) is Hopcroft-Karp, which is basically a specialised case of Dinic's algorithm for max flow.

The maximum bipartite matching is as follows: For each team, if it can possibly use its first option (that is, if no other team starts with the same 3 letters), draw an edge to its first option. Then, always draw an edge to the second option. If you can match all vertices on the left to all vertices on the right, then you have found your solution. Otherwise, there is no solution.

To convert from maximum bipartite matching (say if you have Dinic's bookcode), simply connect the source to all teams with capacity 1, replace all original edges with capacity 1, and add edges from all abbreviations to the sink with capacity 1.

(Unfortunately I had a sneaky bug and failed main tests, but that's the implementation, not the idea.)

A-D: Graphs Graphs Graphs Graphs

B is just tedious implementation, I don't see graphs here.

Well I directly throw some flow models (hope no TLE btw), so it's graph for me :(

If you use Dinic's/Hopcroft-Karp, the complexity should be O(N^(3/2)), so unlikely to TLE.

Funny it can be solved this way. I just looked which teams are forced to have second option (in while loop) and if there are no more than I use first option, that's it. Team is forced to use second option if there is another team with same 3letter prefix (all such conflicts are detected on the beginning) and then in loop I detect conflicts of form that I can't use first option because it is taken.

I used 2-SAT to solve so it's also graph for me.

D is matrix multiplication, I don't see graphs here.

Deleted, sorry.

He's div1!

How to solve D?

EDIT: I'm sorry, it can be found here.

Bonus points for those who will come up with a test to D so that longest possible path has length from interval (10

^{18}, 2^{60}). I think such test will cause many solutions to fail, but maybe such test doesn't exist.It is test 20.

Good job problemsetters :).

i've stuck in B for about 2 hours. What a day!

u r not alone my friend. prestest 3 never cleared for me.

Too. Can't even imagine, where I am wrong. Also, I haven't got any changes in my rating. Is it ok?

I think the system is updating the data.

i think B is Ternary Search problem ..i am not sure i couldn't implement it !

You was right

I did it using binary search :/

Although you failed the main tests.

I commented before systest was over and thats why the

confusedemoji was there. Although, it can be solved using binary search. I don't think I said anything wrong there. My implementation maybe wrong but the idea was correct. :)i solved it using binary search and got ac :)

I used binary search.

how to solve D div1 faster than ?

Using bitset.

I was thinking O(N^3 * log (10^18) / 64), where 64 comes from using OR operator to do something like union of sets. But I didn't have time to code it. Besides, I'm curious to know whether there is something faster than O(N^3 * log (1e18)) too.

Edit: seems like I was beaten to the response.

Any ideas for Div2C?

maximum vertex degree + 1

But what about this test?

5

1 - 2

1 - 3

2 - 4

2 - 5.

your answer is 3 but the main answer is 4!

P.S.oh. sorry! it was my mistake... the solution is correct.the degs are 2 3 1 1 1 so answer is 3 + 1 = 4

Why is my answer 3? Highest degree = 3. Add 1, you get 4

It was just a simple one-pass dfs. For each node, you assign it all values to its children which are not equal to itself and its parent.

Will greedy work for div 2 D?

I pushed every second option in a vector for the corresponding first option, in a map. For this situation, if answer exists, the answer must be the second option, if size of vector > 1. If size of vector is 1, then we can use the first option for each of these clubs, given this doesn't already appear in our answer.

Of course it work, but when you check if there is at least 1 vector with size 1 we have to use the second option, we have to continue searching until all vector of size 1 can use first option

Yeah. My implementation took n^2. I am scared of looking at my submissions page, as I didn't do very well.

That was fast.

Thats the fastest system test ever :p

How to solve Div1 C?

Put in an array the nodes following the visiting order of a dfs on the graph. Then, for each contiguous block of

ceil(2n/k) nodes in the array, assign a robot. If a robot gets no node, just assign any node to it. This works because the dfs spanning tree has exactlyn- 1 edges and you visit each of them 2 times (going down and up); hence, 2(n- 1) + 1 nodes will be visited.Explain to me, please. The total number of verteces, which visited our robots on the basis by K may be <

2*(n-1), isnt it?No. k*(ceil(2*n)/k) >= 2*n

I am sorry. I have to be careful when reading conditions

consider a dfs tree

This is not completely related to the contest itself, but when I go to my room I see a bug in the page title.

It says Room #0 instead of Room #21.

The test data of Div 1 D is weak.

This solution, which consider starting at any city, passed system test. 25265977

If you don't post this comment, I won't know the starting position is fixed forever though I pass it...

So we're waiting for rejudge? Actually, it's weird. So many people were anxious to challenge. For example, test 2 2 2 2 0 2 2 1 should give the answer 0.

Last time there wasn't a rejudge when a Div 2 D (some rounds ago) have quite weak tests though.

P.S. I think my solution does consider starting only at vertex 1 though.

System tests are becoming weaker recently, especially in Div 1.

Bad job problemsetters :(

Omg, I can't believe people are creating anti-unordered_map tests when hacking, but nobody create tests against such thing and people get away with it xd.

When I tried to view the onsite results, I got an error says "You do not have view access to non-public contest".

Edit: It's available to me now.

I tried Div1C with Minimum Spanning Tree and Backtracking. in this solution, searched node of backtracking in MST is no more than 2n-2 then i stored searched node and divided k-set.

...but i got WA. Why?!! T.T

I had exact same problem, can somebody explain what getting WA4 means in terms of solution incorrectness?

Well the answer is somehow I chose a convoluted and wrong order in the DFS to add the vertices. One simply adds the current vertex, and then after every call of dfs adds the current vertex again. Two line change :(

Thank you guys :) was supper easy I don't know how I didn't figure out the answer faster

Pro AF Bug xDContest Submission [TLE in Systest] : 25255834 (How did this shit even pass pretests?)

After Contest [AC] : 25265826

Your codes are not same and have diffs here :

if ((int) colors.size() == neighbours) { break; } "==" instead of ">=" that when there will be a huge vertices with degree of 1 and

ansbe huge do You will get TLE.I know that. I was pointing to the subtlety of the bug, and was surprised as to how it passed the first 47 tests.

Problem B — Spend 1 hours and a half, can't come up with a solution

Problem C — Took me 1 minute to come up with a solution.

Moral of today's contest: read all problems' statement first.

Exact same situation :(

Exactly

differentsituation :(By Technocup onsite results I took C first

Problem C — Spend 1 hours and a half, can't come up with a solution

Problem B — Took me 1 minute to come up with a solution.(Because Binary/Ternary Search is my favorite topic)

Moral of today's contest: read all problem's statement in first come first solve manner.

I got wrong answer on case 16 DIV2 D . Can't find the error . Please help !! http://codeforces.com/contest/782/submission/25264212

me too and it is really making me angry -_-

But my solution still gives the right answer :(( and I still have no idea what's wrong

is it B div2 or B div1 ? http://codeforces.com/problemset/problem/782/B

Div.2 B

can someone give me a simple implementation for B .. i couldn't implement it well ?

I understood solution from my friends code, maybe it can help you.

My friends codehere is my solution 25259459

I use ternary search for finding a meeting place

I used Binary Search on meeting place and keep getting WA at test 24? Could it be Binary Search cannot be used for meeting place?

Meeting place will not always be an integer. So yes, you cannot use Binary Search for the meeting place.

You can use binary search with doubles too...

Oh right, I just realized you can manually terminate it after 'x' iterations similar to how ternary search works.

such a tricky test. Every person stood on the same place. Binary search works out with meeting place. Thank you

My submission: http://codeforces.com/contest/782/submission/25267635

Yes, if you take the limits l = minimum x, r = maximum x you can imagine f(position) begin max(left(position), right(position)). left is the maximum time to get here from the positions to the left and right is from the right. This function is the maximum of an increasing funtion(left) and a decreasing function(right), so it is not always monotonic.

Check your answer when all the persons are at the same point. I also applied binary search on the distances. Since my upper limit and lower limit were equal from the beginning, my code was failing on test 24.

Added an if statement to get AC after the contest.

I did this: Binary search on the answer.

For each value t, you can examine if all of the friends can meet somewhere within t seconds.

First, examine what is the minimum

x_{i}, such thatx_{i}=pos_{i}+speed_{i}*t. Then,min(x_{i}) is the best place to meet if we have t seconds of time. Try to examine if everyone positioned aftermin(x_{i}) can reachmin(x_{i}) after t seconds. If yes, it is possible to meet within t seconds; it is not otherwise.Why we can't see successful submissions?

UPD: Seems it is fixedWhy can't we see the tests ?

Why can't I see other people submissions? All, not just ones from today's competition.

That one statement ruined everything -.-

AC: http://codeforces.com/contest/782/submission/25266644

TLE: http://codeforces.com/contest/782/submission/25257458

I believe we have waited enough for the ratings to be updated :D

what's the solution for Div2D?

Stupid greedy implementation. Try to name the team using the first choice. If you can't, try to name the team using the second choice and rename those that conflict with the new name. If you can't rename those, it means that it is impossible.

Maintain a DSU and place all the teams having same team name [first three characters] in a same component and also maintain the size of each component. Now, if there are two teams u , v (u != v) such that both the teams belong to the same component and the short names formed by using second option for both u and v are equal, then the answer is NO. Now, name all the teams in a component having size > 1 by using second option. Teams having only 1 node in its component has two options, either use first option or second, if both the strings are already used then again the answer is NO, else use either of them which isn't used yet.

PS- This solution is wrong, test data were weak enough to let this solution pass.

Test where it can fail :

ABC DDD

ABD DOG

Maximum Bipartite Matching shall be used to assign the shortnames to the teams having only 1 node in its component.

My solution seems correct (it passed). The algorithm is :-

My actual implementation was O(N^2*log(N)), since I used sets, however with hashing, this will be O(N^2).

This is so because in step 3, atleast one name is pushed to type 2 for every iteration. Giving maximum N iterations => O(N*(complexity of one iteration)).

To administrators and problem designers： I am an OIer from China. (Never asking me why I chose Iceland) Every time I and my friends participated in the round, questions which are written in English worried us greatly. This takes us too much time to understand the meanings . As to myself , in this round , I could have worked C out within 5 more seconds . It takes me too much time to understand B and I could have save the time to work C. So , I request you can prepare questions which are written in Chinese in any form ( a file to download or any else ). Maybe this is not that easy to work , or maybe you haven't the translater , or maybe other countries' participants need questions in their language costs you too much time . It doesn't matter . We wish the problem description can be more clear , terse and easy to understand . This will help us greatly . If you can take my suggestion ， I will be deeply grateful.

Yours sincerely Real_EnderDragon

There are so many interesting things about competitive programming in chinese websites I wish I could be able to translate at least to english... And english is not even my natural language :(

Can some please help me to find bug in my solution for problem B , why i am getting WA in on test 8 . solution Link

I've tried 2-sat in div2-D but got WA. Is enough to test these four cases?

Let option 1 = true, option 2 = false, 'a' and 'b' a team

case 1: first (a) == first (b) ->

(¬a || ¬a) && (¬b || ¬b)— both second optioncase 2: second (a) == second (b) ->

(a || b) && (¬a || ¬b)— a and b must be differentcase 3: first (a) == second (b) ->

(¬a || b)case 4: second (a) == first (b) ->

(a || ¬b)On case 4 check your condition again. I think it must be (second(a) == FIRST(b)).

Ops, you're right, I made a mistake in the description but it's correct in the code, thank you! Any other observation?

In case 2, when first(a) != first(b), you can choose option1 for both so it should be only

(a||b).That was it, it worked now, thanks a lot!

waiting and waiting for rate change ...

Is the ceiling function notation used in the statement of div1.C considered to be general knowledge? I've never seen it before and got a WA on system tests because I thought it was an integer function.

It would be nice if the statement always contained definitions of all nontrivial functions used in it (IIRC, they even provide a definition of bitwise operations in TopCoder SRMs. I don't think it hurts anyone).

I think it's quite common. I've seen it at least in

`Introduction to Algorithms`

by Cormen and others and also in`Concrete Mathematics`

by Knuth and others.But agree, that statements should contain definitions. Or maybe we can have some extra page

`Notation`

on codeforces like some of the books have, that will cover common cases.F looks moderately interesting, and there has been no discussion on it, so maybe I'll start with an idea.

We binary search on the answer D. Hence, we only need concern ourselves with whether it is possible to achieve synchronization with distance D.

Consider the distance between buses 1 and 2 as they around the circuit. Consider a time segment through which neither bus turns a corner. Then the distance function over this segment only has at most 1 turning point, and so will pass D at most twice.

In fact, we know if there is a turning point in this segment: the only difficult case is when the buses are travelling perpendicular to each other, in which case, pretending that the line segments extend infinitely into the past and future, the turning point is precisely the point where the buses are equidistant from the intersection of their lines, with one bus moving towards and one bus away from the intersection. This is the only possible turning point of the function.

Hence, with some effort, we can find the points where f(x) = D, and hence the intervals where f(x)>D. (I'm fuzzy on this front, but binary division seems like it'll run in time.)

Assume we can compute the intervals on which the distance between the 2 buses exceeds D. Then, we can take the union of all these intervals modulo the distance between the 2 buses, ie. treat the time as cyclic with period of (length of route/number of buses). If the entire interval is covered, then D (or at least, D-1e-7) cannot be achieved. Otherwise, D can be achieved, and we can proceed with the binary search.

Checking of unions of interval over a line segment can be done using a set, for example.

Any errors/simplifications?

That's more-or-less what I was implementing, but couldn't find the bugs in time. For each interval where the two buses are moving in a straight line, the vector between them varies linearly, and asking for the two times at which the distance exactly equals D requires solving a quadratic equation. If the determinant is negative then they are never within D of each other, otherwise you get an interval, which then also needs to be clipped to the time interval you're considering.

I took intersections of intervals when the distance is at most D rather than unions of the complement, but it's equivalent. It also simplifies things to insert breaks at every multiple of the inter-bus separation (i.e. treat those times as if bus 1 made a turn), so that the intervals you get never wrap around.

Div2 B using while(r-l) > 1e-7 failed system test (time limit) http://codeforces.com/contest/782/submission/25253014

using while(r-l) > 1e-6 after the contest accepted http://codeforces.com/contest/782/submission/25265567

life is so cruel

I used

`1e-9`

in my submission, 100ms. Life is so kind to me then!i think because you used binary search, you always divide by 2 so you will not face numbers like 0.3333(with infinite threes), but i used ternary search (has dividing by three) which will face this kind of numbers, so it will be truncated then due to the double precision and a high probability that r and l will reach a point that they will not change after it in the loop (so infinite loop occurs)

this gives me a lesson in the future to include additional check to check that if the calculated numbers (boundaries) in the loop didn't change from the last iteration then break from the loop.

I also had an infinite loop in ternary search, but my lesson is a bit different. Instead of checking absolute error, I'll just do a loop with a fixed number of iterations:

I think Dataset for Div2 D is weak. Say for this case

This submission outputs NO

And The Winner's submission outputs YES.

Hi guys, can you explain me, why this Solution has TLE? Sol when I did another using Boolean tab it has compiled program in 140ms, is this really that big difference?

Of course it does. :)

Your algorithm implemented with a list is O(N^2) for n = 10^5, this will definitely give a time limit.

The other one is O(N) which passes. If that confuses you read about big oh notation, and then see for what input size you can do each one.

Div2 D has a very weak test cases

some ACC soultion output YES and some output NO in this case

3

AAA BBB

AAA CCC

AAD BBB

I used ternary search finding derivative tending to zero to solve Div2B 25253744

As F(position)=time has only one local min, then it is the global one so the only point with derivative=0 is the solution (and both left and right parts of the graph are infinite descending and ascending)

Thank you calculus :)

Someone with WA16 on Div2D/ Div1B? 25271010

I know its too late to respond, but i am stuck right there.

Could you please help??

How can I implement Div2 B using binary search ?

Let the function

`f(x)`

return the time to gather to the point x. The domain is [a,b].`a`

is the leftmost coordinate and`b`

the rightmost one. It turns out that f is first decreasing and then increasing after some`X`

. The task is to find`X`

, at which`f`

attains its minimun.Take the middle point of the current interval.

Compute f(mid-eps), f(mid), f(mid+eps). Take eps a very small positive number like 0.0000001.

If

`f(mid-eps) < f(mid) < f(mid+eps),`

then the answer is in`[a,mid]`

.If

`f(mid-eps) > f(mid) > f(mid+eps)`

, then the answer is in`[mid,b]`

.Else the answer is

`mid`

itself.Just simply implement with a while loop.

I have done exactly that 25253744. Only that I called that difference derivative and my goal was to find when derivative was 0.

Good explanation. I managed to get AC based on this. I have some complementary questions though.

When you solved this problem on the contest, did you find exact proof for the fact that "f is first decreasing and then increasing after some X"? Or just viewed some example data and presumed that this will hold for all input?

How to choose proper epsilon? If I choose too big than relative error will be big also. If I choose too small than maybe I can't get differences in the value of the function. Is it possible to bound the relative error in the value of the function based on epsilon. Is there any exact proof?

I just found it intuitive that while you are getting far from the optimal point the time is only increasing. I don't have any idea about the rigorous proof. Regarding the epsilon I usually take it a bit smaller than the relative error.

I think in those cases fixed number of iterations is more predictable, more safe, and more practical. Fix on a number of iterations enough big to avoid precision problems, and enough small to avoid TL. usually around 100 satisfies both with margin

My approach is a little bit different from the other one.

Do binary search for the point in the interval. For a given point, there will be a friend from the north who reaches the point last of all friends from the north, and similarly a friend from the south. If these two are equally slow, you found the answer. Else move towards the slower one.

Binary search (on doubles) for the minimum time

tit takes for all the people to meet.For a given time

t, note that the set of places personican reach within timetis the range [pos[i] -v[i]·t,pos[i] +v[i]·t] on the number line. It is possible for all of the people to meet in a given timetif there exists a point inside all the ranges, or if the intersection of all these ranges is not degenerate. To check if the intersection of all these ranges is not degenerate, we can letlbe the maximum of the left endpoints of these ranges andrbe the minimum of the right endpoints of these ranges. Ifl≤r, it means the intersection of the ranges is not degenerate, while ifl>r, the intersection of the ranges is degenerate.Code: http://codeforces.com/contest/782/submission/25277237

How to solve Div1E?

If you can find quickly, given where each ball starts its fall, which platform it lands on, then the rest of the question is easy. There are at most 3e5 such computations to make.

Consider processing each of the 3e5 queries from left to right (each query asks, for a starting point, where it ends). For each platform i, there is a maximum height h_i for which it will accept. Consider an array storing h_i's. Then, if we have a ball (index i in array) falling from height H, we simply want to find the minimum index greater than i of the array with entry at least H.

We just need to support queries of the form: i) Point update ii) Point update iii) Range query of maximum in range

This can be done using a segment tree. Binary searching to find the minimum such index gives O(lg^2 N) every query, but can be sped up to O(lg N) if we build it into the segment tree.

I'm wandering if there will be an official tutorial. Or if anybody could tell me that if is the best time complex for Div1.D? Thanks a lot!

I think many solution have that complexity as well. (at least mine is)

not rigorous DATA FOR YESTERDAY'S 781B Innokenty and a Football League THAT SOMEONE USE WRONG PROGRAM TO PASS THE TEST. WE MAY INSERT TWO DATAS THAT CAN SOLVE IT INPUT 1 3 ABC GGG ABG GCC ABD EEE OUTPUT 1 YES ABC ABG ABD INPUT 2 3 ABC DDD ABD CCC ABD EEE OUTPUT 2 YES 3 ABD ABC ABE

How to solve Div1D/Div2F?

Let me define it:

f(0,0) = P;

f(0,1) = B;

f(1,0) = f(0,0) + f(0,1);

f(1,0) = f(0,1) + f(0,0);

f(i,0) = f(i-1,0) +f(i-1,1);

f(i,1) = f(i-1,1)+ f(i-1,0);

...

Let DP[u][j][x]= (the set of vertices that the vertex u, on the j-th, x-th state of this function can reach);

Basically, for each edge u->v,w , you must add: Dp[u][0][w][v]=1; then, for each i until (60) for each vertex u for each i, w of the funcion f , you must calculate the Dp[u][i][w];

Remember that we have calculated the Dp[u][i-1][w]! Then , for each vertex v , that you can reach on the state(u,i-1,w) , we have to add the set of vertex that the state(v,i-1,!w).

to implement union of sets efficiently you could use a bitset!

the complexity: O(n^3*log(10^18)).

I hope it helps ! Sorry for my bad english!

What is the function f for?

I think

f(I, 0) is the correct sequence obtained after (I+ 1)^{th}step andf(I, 1) is the sequence obtained by changing allP'stoBand allB'stoPinf(I, 0).Yeah..thanks. Makes sense now.

yes! it is !

Hi, which is bitset average/worst complexity?

Can someone find the mistake in my DIV1 B/DIV2 D , it fails on test case 16 http://codeforces.com/contest/782/submission/25261134

It is very difficult to figure out exact problem but I also got WA in pretest for this case first then I corrected it with some small changes in the code. I guess you must be making similar mistake. The second loop that you are using, I guess it should be used n times rather than only one time. The problem is when you make someone switch from first to second option this may create some new conflicts which you have to correct in next iteration. One more thing I can see is you are using the array taken which I think is not necessary, You can just update the map every time (In your second loop). If you still have some problem, Have a look at my solution it is quite similar to yours.

thanks for the reply, " The problem is when you make someone switch from first to second option this may create some new conflicts which you have to correct in next iteration."

can u give an example for such situation ?

1 ABB ABD

2 ABB ABE

3 ABX ABY

4 ABF ABX

5 ABE ABF

Have a look at this. The first column is first choice and second column is second choice. I guess this is what I am talking.

Why loop N times? How does iterating N times completely remove all conflicts?

In 1 loop you at least fix one of the relation. That relation you will never change again. So after n loops all of them will get fixed value. ( Another intuition to this is it is like we are finding a matching using Ford fulkersons and finding the augmenting path in every loop, So after finding n augmenting path, We find a matching)

Did you find a solution? I'm also struggling with WA on test #16 http://codeforces.com/contest/782/submission/25303767. Did you find a have a simpler failing testset? On:

5 ABB D ABB E ABX Y ABF X ABE F

I get (correctly):

YES ABD ABE ABY ABX ABF

EDIT: Oh, forget it, think I found it, I go wrong on:

4 AAA B AAA D AAC E AAD C

In DIV2 B, I get TLE on test 3 http://codeforces.com/contest/782/submission/25283174

but if i write functions for min and max , it gets AC http://codeforces.com/contest/782/submission/25283217

why is it so ?

Your mistake —

`double d = 1e-9`

. If you write, for example,`double d = 1e-7`

you will get AC with both min/max functions. Also you can use`long double`

, it (probably) takes 16 bytes.Double numbers on server (probably) takes 8 bytes, so maximum count of significant decimal digits is about 17.

Sorry for my english:)

Will there be an editorial? Or is there anybody can tell me why simple dfs can work on DIV2-E/DIV1-C and how to prove the correctness? Thanks a lot

Basically the idea is that if you solved the problem for

k= 1, then you can solve fork> 1 as well since you just take the solution fork= 1, split it intokequal parts and make sure each part is nonempty.To solve for

k= 1, you need to go through all vertices in the graph in at most 2nmoves. Since the graph is connected, we can find a spanning tree of it. Now, we can just do an euler tour on the spanning tree which contains 2n- 1 vertices. DFS can be used to find the spanning tree.How quick! I get it, thank you.

For div2/B is there any solution rather than binary/ternary search?

This is the first time to go to blue Expert! Incredibly!!!

And this was my 7th time to become Expert... I wish I could be that excited. :)

I just use codeforces one year! :)

You dropped down to Expert, I rose to there.... Perhaps next contest will kick my ass dropping down to specialist! ahaha

Edutorial please...

http://codeforces.com/blog/entry/50854

Is there an editorial for the contest KAN ?

Here this is http://codeforces.com/blog/entry/50854

In div2D, I want to know the answer of below case. 3 ABC DEF ABD EFG ABE DEF

I think the answer is YES(ABC,ABD,ABE), but my source print "NO" although I got an AC.

my answer is Yes !

For people who are having trouble with Test case 16 on DIV2 D, consider this test case:

4

ABD BCD

ABC DDD

ABR CDD

ABR XYY

The answer should be:

YES

ABB

ABD

ABC

ABX

My solution which gives WA on test 16 fails this test too.

Solution for Div 1 E ? Anyone ?

can someone please find why i am getting runtime error in python2.7 in problem C div 2?

Here is my code, a big thanks in advance