Hi, Codeforces!

At 13.01.2019 17:35 (Московское время) will start Codeforces Round #532 (Div. 2). Round will be rated for second division (rating below 2100). As usual, participants from the first division can participate in a contest out of competition.

The round will consist of 6 problems and you will be given two hours to solve them.

Task prepared by me, den2204.

Thanks MikeMirzayanov for the platform Codeforces, KAN for round coordination and help with preparation. Thanks gritukan, Andreikkaa, flyrise, Um_nik, Aleks5d and osaaateiasavtnl. for help with preparation and testing round, arsor for translating statements into English, arsijo for assistance in conducting the round. Thanks Andreikkaa for the idea of the problem E.

Good luck!

P.S. It is my first round on Codeforces, please, do not judge strictly :)

**UPD:** Added editorial. Thank you all for participating and for the feedback on the contest. I apologize for the leap in complexity between tasks and for weak pretests for some tasks, I’ll take this into account when further preparing contests.

Auto comment: topic has been updated by den2204 (previous revision, new revision, compare).Honestly, not bad for your first contest! I hope you had a great time creating the contest and all of us participants would like to thank you for spending your time on this! We all really appreciate it! Hope to see you again as a writer den2204!

And to all participants: Good luck and have fun at the contest tomorrow! <3

Segment Tree in E???

may be fft

Heh. den2204 changed it to:

Thanks Andreikkaa for the idea of the problem:PE.hope it will be an easy round with the last question not exceeding the difficulty of 2200

Finally a contest without faked colors !

:D

But isn't this an usual start time for codeforces rounds?

:d

den2204, congratulations for your first contest as a problem-setter...

This is going to be my first contest of the new year. I am so excited. xD

I didn't find you in the contest today. You got paralyzed with too much excitement or something?

Couldn't give it because of some urgent work. Thanks for noticing though.

Obviously my classmate after all xD

good!!

:d

Congratulate with your first contest, I think it will be greate :) Good luck all :D

Please make strong test cases

Lol. Tera hack ho gaya tha na B kal

I want a good contest for everyone

When there is a new problem setter its usually 2 condition : 1) the tasks are creative and fun to solve 2) the tasks are time taking and not fun to solve I hope for (1) :)

I hope too :)

Both condition are satisfied

i hope your first contest will be great ^^

Scoring distribution?

why geometry??

why not?

Rather than complaining about it, maybe it would be more beneficial to get better at the topic, because it's not going away.

why not?because it is completely pointless and rather than wasting time "getting good at it" for no reason, they might as well give us better and more useful problems overall.

Do you have any reason why other problems are “better” and “more useful” besides just your opinion?

It’s not pointless if your goal is to get better at cp. You’ll end up doing better in contests because you’ll know the topic.

And if you don’t want to learn geometry either, that’s perfectly fine. Just don’t start complaining about the questions when you can’t do it, since you made the choice not to learn. There’s 5 other questions for you to try.

I'm neither with any sides nor saying geometry is bad, but you know, memorizing( or even searching ) a formula and coding an O(1) program is not really fun.

All you need to memorize is the definition of sin function. If you can't memorize 8-th grade math, then I have bad news for you.

How does the definition of sin function help in solving this problem?

You have to connect center of inner circle, center of outer circle and point of tangency between two outer circles, In this triangle we know two sides, one acute angle, and also we know that angle between tangent and radius is right.

I'm not saying that this problem is innovative or interesting, but in no way that's because it is geometry or math in general.

OK, I'm too stupid, and I know it!!!

BTW in our country they teach it in 11th grade, and I haven't reached there yet.

And again; BTW I know what sin func is.( But still stupid )

in your country you learn it on the second lesson of the 10th grade math!

Your school is very ghavi!

jfyi Iran is quite good at math:

http://imo-official.org/country_team_r.aspx?code=IRN

Good luck studying more advanced mathematical concepts like tanh or derivatives.

exactly, it isn't fun nor useful.

It can be boring if you can instantly come up with the correct formula and code it up (but it's also good, since it will give you more time to work on other problems).

If you don't know the formula, you can spend some time deriving the solution using what geometry knowledge that you know.

And isn't going through the process to reach a solution the fun in cp?

so in conclusion,

it is pointless. period. glad we could all agree.Sorry if you didn’t see it before.

i saw it but because you have anime profile picture i have automatically skipped it so i didn't count it.

What is

useful?So geometry is useless, but wasting time to create fake accounts is useful?

how can an account be fake?

You didn't have the balls to post that comment with your actual account because you feared downvotes, so you created this one. Do you honestly think we don't know that?

THISis my main account. you can ask the codeforces administration about this.your downvotes don't affect me, they are worthless because i know what i say is true.

You yet to do a contest in your "main account" and you are complaining about problemset. WOW!!!

? what are you talking about?

Before trying to make the community fool, you may not notice your color is still black now :P :P And you may be forget people can go to your profile and see the number of contest you attend till now.

you imbecile just because it glitched for you doesn't mean that's how it is for everyone.

Illuminati wants to know your location.

There is a drastic difference between A-C and D-F, so the contest turned into speed coding.

True. but problem D was not that tough. But don't know why people couldn't solve it.

“Thanks to MikeMirzayanov for the platform,…” Standard Reply. Cheers for Codeforces! (big :D)

I hope you not to repeat it

That's not constructive criticism.

I am really sorry but , the contest went into wrong direction and I hope the next contests that prepared by him to be more good and avoding the mistakes that has happend at this contest . hope good luck for all .

you’re right it was trash

Was that your another fake ID or something else?

?

Worst Problem statement with worst Explanation. Wasted 10 min in just understanding the problem Statements

Also boring A,B,C and unbalanced problemset. Slight error in implementation and you are very low in ranklist.

That happened to me :(

i am still trying to understand D. LOL

....

gg

Why D, E, F so hard? Massive AC-gap between C & D :(

For me E is easier than C.

how to solve E. can topolozical sorting be applies on cyclic graph?

That's right.But D is too hard to put in a Div.2 D I think.

C was pretty straight forward. If you know Trigonometry, It is easier than A , else you wouldn't be able to do it :( How about E ? Didn't read that. Was it a data structure problem ? Or something else ?

A, B, C = A and D, E, F != DIV 2.

In problem F, are we supposed to answer the queries for maximum subarray xor? Probably somehow offline?

No, we have to choose some numbers from l to r s.t. their xor is maximum ... It's a problem similar to https://www.spoj.com/problems/XMAX/.

I am aware of that, let me reformulate my question. Is that problem equivalent to finding maximum subarray xor on segments of the given array?

No, it's related to finding independent basis vectors of l to r. It can be done using gaussian elimination. But my solution was N*logN*20^2 which is not enough to pass t.l.

In general, maximum subset xor and maximum subarray xor aren't going to be the same — subarray is a fairly restrictive condition when you need to pick things.

For example, take

`a = [5, 6, 2]`

Maximum subset xor is 7, maximum subarray xor is 6.

Isn't subarray a subset of an array?

If that's the definition you're going with, then yes, the answer is maximum subarray xor.

I'm going with the more commonly used definition for competitive coding, where a subset is your usual subset and a subarray is a contiguous segment of the array. Similar to how a substring is a contiguous segment of a string.

Something like this, basically.

Can it be done without Gaussian elimination?

Maximum subarray xor can be done with a trie — see here for a tutorial.

Maximum subset xor, I don't think so. I'm not aware of any other method at the very least.

I found another link that answers this. https://math.stackexchange.com/questions/48682/maximization-with-xor-operator

How to solve D?

Bring the king to the center. Then count the number of rooks in each quadrant, and go diagonally in the direction opposite to the quadrant with the least number of rooks.

Verify as an exercise that this works.

How to solve F?

Take yesterday's G -> TLE with seg tree -> use divide and conquer that you can use just prefix/suffix -> done

Yeah I got TLE with segment tree, very nice solution, thanks!

Is it possible to merge two independent vector sets in better than 20^2 time?

I don't think so. My solution computes suffix/prefix when dividing [l, mid) [mid, r) (to add a number to a prefix/suffix is O(bits) time) and then solves the queries with l < mid and r >= mid in O(bits^2).

Your Gauss struct is black box for me.

If you can provide some intuition how those

`max and min`

are working?First time I have seen inserting an element in O(bits).

My implementation in O(bits^2) was giving tle.

You can add number by number. So instead of thinking of a N*BITS matrix when you have N numbers, think of a BITS*BITS matrix where you insert one value at a time. Also, don't use vectors in code like this that uses a ton of memory, it'll use a lot more memory and it'll have more cache misses. For max, you have the basis so just go from the number with highest bit to lowest bit and activate the bit whenever you can.

Got it. Thank You.

What I have understood is that each no in your table has a bit (the most significant bit) associated with it. Similar to the basis vector which has at least one dimension with them.

And that no is responsible for activating(max)/ deactivating(min) that particular bit.

P.S. After reading I got core idea of your implementation. But was bit confused with

`just go from the number with highest bit to lowest bit`

So, I added prt function in your implementation. And then I came up with the above explanation.Thank you. :)

In your submission (linked below), the

`Gauss::add()`

function confuses me.I think that the invariant for the

`Gauss`

class is that if`table[i] > 0`

then the $i$th bit of that needs to be set. But if you don't make that check in your`add()`

, this invariant could be violated.Your submission did get AC, so what am I missing here?

48339644

Edit: True, let me think how that worked rofl. So... I got careless and think that code shouldn't work, systests were weak?

You can split your array in chunks of size 20 and use sparse table on array of size n/20. Query can now be answered online in 20*20 time by sparse table query + adding elements partially belonging to a chunk one by one.

Nice! I thought of sparse table but I didn't think of dividing into blocks, that's a cool idea.

Does E have the following solution?

Sort all the edges according to

`c`

. Binary search over this to find the edge so that no cycle is present. Compute the topological order of this graph without cycle and add reverse remaining edges which violate the order.Sorry if I understand your idea wrong, but don't you mean to reverse all the edges that have a lower cost than the middle in your binary search ? I thought the same but there might be some edges that you can reverse that should not be reversed which made it so much harder

You cannot reverse all the edges which have lower weight because you might end up getting cycles in new graph. Instead what you do is reverse those edges which violate the topological order as a directed graph has no cycles iff it has a topological order.

The problems were actually pretty good, but the difficulty gap was pretty bad. Hope you can balance the problems in the future!

How we were supposed to locally test D?

Write your own local judge, or don't write with mistakes.

Anyways, if you submitted and get a wrong answer on the first test, it will tell you what went wrong, for example "invalid move for king", which is still helpful.

I scaled down the problem by a factor of 111, i.e. 9x9 board and 6 rooks in play.

That's a really clever line of thought.

I can see that 1 dimensional properties and 2 dimensional properties of the problem will remain the same.

But maybe, just maybe, the properties of the problem that require an interaction of a 1D vs 2D would not hold.I'm sorry that I'm very vague, but I'm not very clear about it myself. I also can't think of an example of such a property.

How to solve E? I thought of using finding a maximum spanning tree and every not used edge would be a flipped edge, but I didn't know how to implement a dsu for directed graphs.

binary search the answer, then the problem becomes to check whether it has cycle.

Thanks

Hey, one quick question, would we reverse all the edges whose c is less than or equal to mid?

No, find a valid topological ordering for the graph with > mid. Now, just orient the edges <= mid according to this order.

I don't see why I can ignore edges to check of cycles just because I can choose the orientation of them on the directed graph.

Given that you ignore them, you have a topological ordering. Now, you can direct them using the topological order (from the vertex that comes first in the order to the one that comes second).

Thank you!

can you explain what u mean by bs over the answer? what answer

When a problem is asking for the minimum

ksatisfies a conditionT, andksatisfiesT=> anyk' >ksatisfiesT, and for a numberk' you can check whetherk' satisfiesT.You can binary search

k. IfksatisfiesTtryksmaller, else tryklarger.ok skylinebaby i understand .

regarding this question everytime we should choose a k using binary search.

then we can build a new graph with edges > k , if this graph forms a cycle , then answer will obviously will > k , and if not then answer will be <=k.

this way we can get the best k .

now , how to proceed further . which edges to revert so that graph becomes acyclic .

You can topological sort the graph with edge >

k, because the problem don't need to minimize the number of edge which should be reversed, you can simply reverse every edge which has the wrong direction against the topological sort.wow.. thanks great idea.. thanks for help!

so is it reduced to can we make topological order of edges > mid, because if we can do so we can ALWAYS adjust remaining edges? . please help, i m not finding the way that how remaining edges can cause problem to topological order for why we need to orient remaining edges to topological order

Yes, if we can make the topological order of edges >

k, we can always adjust remaining edges < =k, because we can just follow the topological order. On the other hand, if we can't make the topological order of edges >k, that means there is a cycle with edge all >k, and we can't satisfy the original condition by reversing those edge < =k.oh thanks bro, just got it accepted :D

Wasted the whole time on C. Dunno why got WA on D. Coded E just after the contest ended. Gonna kill myself if it passes.

So, are you dead?

Honestly speaking, the gap between C and D is very large. C is the normal Div.2 B and B is the normal Div.2 A.

Surely you didn't send problems to the problem pool like what Arkady did on the problem B. I mean it was like, n=6, m = 6 and 112556 then the answer is 000001

Need solution on D too. It really looks like pigeonhole, but I don't know how to do it.

D and E were easier than C

Well, depends on perspective IMO.

For me, I think C is easier than D and E. You just need to write some geometric formula and derive it.

Well, actually I can't tell much because I didn't finish D and E.

Totally awesome!

C solution

Geometry problem ---> T̶r̶y̶ ̶t̶o̶ ̶s̶o̶l̶v̶e̶ ̶i̶t̶ google it.

What were the hacks for B and A? I managed to hack someone on B because his implementation had a time complexity of O((m — n) * n). Are there any other hacks for B?

I passed pretests on A but failed system tests by thinking that you could opt to not remove any tabs. The case 3 2 1 1 1 will break this

How to solve C guys? I could not come with a solution during the contest :(

Solve your homework on triangular formulas. If you did not take them yet, well you are rekt.

The centers of the n outer circles make a regular polygon with n sides. Now focus on the centers of 2 adjacent outer circles and the middle circle. You have a triangle with side lengths (r+R), (r+R) and (R+R). You can also find all the angles in the triangle. Finally, you can use cosine theorem to get a formula: (2*R)^2 = 2*(R+r)^2 — 2*(R+r)^2*cos(A) [A is the center angle, actually A = 360/n]. Then, you can simplify and use the quadratic formula, or use binary search.

What I did was visualized that the centers of the outer circles when connected formed a regular polygon with n sides. That polygon has its center coinciding with the center of the inner circle. We can calculate the length from the vertex of the polygon to the center of the inner circle using trigonometry and hence there is only one variable i.e R (here side of the polygon will be 2R). I am almost sure of my solution, I didn't take time to prove what I conjectured, so will come to know exactly if my hypothesis is right when final results are out.

This is my idea,may be inefficient.

draw some cases,we can find:

The center of inner circle can make a Isosceles triangle with two conterminous outer circles' center,and two angles will be ,difine this as θ

then we have (

R+r)^{2}=R^{2}+ (R·tan(θ))^{2}If you connect the centers of two adjacent little circles and the center of the big one, you'll get a triangle. The sides of this triangle have lengths r+R,r+R and 2r

. A little trigonometry will get you that the top angle is

θ=2arcsin(r/r+R).

Since you want the small circles to form a closed ring around the big circle, this angle should enter an integer amount of times in 360° (or 2π

if you work in radians). Thus,

θ=360°/n.

From this, you can compute that

r=Rsin(180°/n)/1−sin(180°/n).

All you gotta realize is that centers of outside circles must be 2R away from each other. How to get centers? Well just do x = (r+R)cos(t) and y = (r+R)sin(t), where t = (2pi/n)*i. If you want to visualize draw two circles tangent to each other with radius r and R. They will be r+R away from each other, and assuming one of them is at 0,0 the other's coordinates can be gotten from sin and cos.

Binary search: if distance < 2R the circles are too big otherwise they are too small.

Please tell me how to debug intractive problem or what is pretets 2 for D. Idea for D: Place king in 500,500. Divide chessboard in 4 pieces. Each piece has one corner as king position and one corner as chessboard corner. Find which of this 4 pieces has smallest number of rooks. Go to opposite corner with king moving only diagonally. Convince yourself it always works.

This is correct and actually needs at most 1000, not 2000 steps! Really cool problem.

I debugged on 9 x 9 grid with 6 pieces.

Nice idea.

What I was trying to do was sort the rows and columns (combined in a single container) based on the number of rook each had. And then try to move my king towards the row or column having the most rooks. This is probably wrong (or take more than 2000 steps) because in one step the king changes either a row or a column. But in you case it changes both x and y thus a better chance of quickly finding a solution.

Unfortunately, one must check that the king doesn't move to a space already occupied by a rook (which I failed to do). This is probably where the factor of two comes from.

If we try to move to a cell with a rook, it means we are trying to move diagonally! Otherwise the king was already hit. Therefore, we can just move either horizontally or vertically to a cell adjacent to this rook. And then we are hit and the game ends. So the number of moves will really be at most 1000.

I did it as well. Can anyone take a look and tell me what is wrong with this? Even passed pre1. 48356184

I did the same with WA6. I think what matters is how you get to the corner. You have to try to move diagonally as much as possible. After considering this, got AC!

Thanks for reply but, already found mistake it was checking if(x!=500 and y!=500) instead of (x!=500 or y!=500) ;)

I'm surely misinterpreting the problem/making some really silly mistake, but I think the king can't guarantee having a check ?

Define

S_{i}to be the configuration where the black rooks are at the cells {(1, 1), (1, 2), ..., (1,i- 1), (1,i+ 334), (1,i+ 335), ..., (1, 999)}. Initially start withS_{1}and the white king at (1, 1). Now at some step, if the king'sxcoordinate changes fromxtoy(with |x-y| ≤ 1) then it's easy to see we can move some rook so that the rooks configuration become , so the king can never guarantee a check ?But in your example all the pieces share the row, so the king is already hit by all the rooks.

Ooops thanks, it was embrassing.

Just curious to know, why does this submission on problem B doesn't get runttime error when he declares array of size exactly 100000.

My hack test

Why would it get runtime error? n, m <= 100000 so it seems fine.

array size has been declared as 100000. hence, array element range is between 0-99999. but he does

`arr[aa]++;`

hence it should get runtime when aa = 100000. Sorry, correct me if I am wrong.

Ah, so he is accessing an element outside of the array. If he'd used std::vector instead of a plain array he would have gotten a Runtime Error. However, since he's using a plain array, his solution is doing

undefined behaviour, which is bad, but most often it won't be a problem because he's accessing memory right next to the array.When you declare an array.Usually you will get an array a little bigger than you declare.That works in most computers.Include my computer,Codeforces computer and our school computer.And even a[-1] won't get Runtime Error in some computers.

Why u kill ma brother roflsroyce and lmaoburghini and mute ma friend keksla. I have ponderd upon the ratability of this contest

OMG! I had a clean solution for D, but didn't consider the case of eating a rook. Didn't look at the CF error message :(( It was saying

king ate rookTrash contest...

1500 point problem solved by 3.5k and D problem solved by only 125 people. Also E problem is just classic know problem added binary search(here is the solution). I think F also range version of the XOR max problem(here is solution).

If you want to do such contest, better to not do. Educational Rounds much better than this kind of contests...

Problem D was not actually that difficult. The entire contest was consistently simple IMO, which is fair.

I do agree that this contest is more like an Educational round, tho. Because it's very easy to google solutions to some problems.

I do not want to judge you strictly. But What the hell.

I understand that it is not easy to make a well balanced contest, but the big jumps in difficulty of problems are really obvious in most of recent codeforces contests.

how to solve B?? Can anyone please explain??

I got pretest pass by maintaining a set {1...N} and erase element x when I see x. If I saw x after I erased it, store that information. When set is empty, re-construct the set, without the elements I stored information about before.

For example, 3 9 1 1 2 3 3 3 2 1 3

Set : 1 2 3 (initial)-> 2 3 -> 2 3, (1 seen once more) , 3 (1 seen once more) -> X (first Round), Reconstruct only 2, 3 -> 2 -> 2, (3 seen once more) .....

maybe you will get TLE set is too slow gratus907

Isn't set insert and erase both O(log n) operation?

It was clear that whenever i get all type of elements atleast multiple of n times then i can create a contest. What I did was maintained a frequency array freq for each type of element. As soon as i get element of type X i increased it's frequency say new frequency is z. Also i maintained an array for keeping count of how many elements i have of particular frequency z. If i have n such elements that means i can create a contest so i incremented answer by one.

Is it possible to hack someone's solution after the contest? Or how can I practice hacking (besides educational rounds)?

recurze, your B solution will probably run out of time on a test like:

I tried to hack it during the last 2 minutes of the contest but forgot to put an endl after the 1st line, so it wasn't accepted.

My the very first hack :D

I was more concerned about the inner loop but figured it'll only run (m/n) * n times so all good but didn't realize the if condition would run O(n). I tried to port my old solution using sets to using frequency array and didn't think much.

My own solution got TLE on test 41, lmao.

I guess, using std::set was the best thing to do.

Any structure with support for key-value pair (i, frequency[i]) would do the trick I suppose.

Codeforces turning into Codechef :P

How to solve E?Here's my idea: Run a DFS from any node, let onstack[i] denote if the ith node is currently being processed, also intime[i] denote the scan time of ith node(similar to topological sort algo). Now maintain a segtree for onstack nodes only, and use RMQ query where index is the intime of nodes. Whenever we find a backedge we calculate the minimum weight edge in the cycle using the segtree. Is this idea correct? Is there an easier way?

I'm not sure if this is easier, but this was my solution.

## Task E Statement:

Given a weighted directed graph, convert it into a DAG by reversing edges. You have to minimize the maximum weight of the edges that you reverse.

If the given graph

G= (V,E) is already a DAG, then the answer is 0.Let us choose a value

karbitarily. Let us call an edge heavy if it's weight is greater than k, else it is light.If

G_{heavy}= (V,E_{heavy}) forms a DAG, then we can find a topological sorting ofVinG_{heavy}. We can then arrange edges inE_{light}to be consistent with the topological sort.We can find the minimum value of

kso thatG_{heavy}doesn't have cycles using binary search.Are you reversing the direction of all the light edges? If so, then there might be a case where the original direction of a light edge might be a part of one cycle and after reversing it, it becomes part of other cycle.

No, not all the light edges. We find a topological sort of

G_{heavy}. Then we can ensure that all the light edges are oriented so that they are consistent with that order.We will only reverse those edges that are inconsistent with the order.

This is my submission, but I'm not sure how readable it is. I don't use the terms heavy and light in the code. It also hasn't passed system tests

yet.48354300

EDIT: Passed system tests.

what if the selected edges created 2 graphs, now we have 2 topological sorts what if we need to connect node from the first graph to another node from the second graph. example

now we need to connect nodes (D, E) but order[D]<order[E] so we reverse it and we add it to the answer, but if we didn't reverse it the graph is still DAG. how we handle this case, it seems I'm missing something.

You mean a disjoint DAG? That's a DAG as well, and it has a topological sort as well.

In your example, there are many valid topological orders. Let us choose any one, say "EFGHABCD".

If we have edge between D to E, then we must have the direction as E--->D (because that is what the topological order tell us).

If the given edge is D--->E, then we will reverse it, else we won't.

Hope it is clear now.

Thanks for your respond, yea I understand that. but my question is, In the example, we can generate 2 topological sort

let's say we have edge E->D if we selected the first topological sort then we need to reverse the edge if we selected the second topological sort then we don't need to reverse the edge. so I see that selecting the topological sort actually affect the results, but it's not clear for me why selecting arbitrary topological sort is ok, did you got what I mean?! :D

I think I understand your question.

The observation is correct that choosing a specific topological sort affects the final resulting Graph (after reversing edges).

The claim is that we could use any of it. But once we choose a topological sort, we can't change that.

I think the best way for you to get this intuition but by generating examples.

Sorry if this is obvious but how is minimizing the maximum weight of reversed edges the same as minimizing the sum of the weights of the edges we reverse.

We aren't minimizing the sum.

I can't find this quote in my comment.

Trash. Again.

It's den2204's first time creating a contest. Cut him some slack!

Pretests of E are weak. I did not get TLE for the following DFS routine:

Good round! Thanks for problem D!

In these cantatas, B has changed with C and D with E; and I think that it not be good;

I think so

contestant.

If I submitted two solution for same problem and both are right then which will be judged first or second one?

second. first will be skipped.

I think only the last solution you've submitted that passes the pretests will be taken into account.

The last solution that passed pretests will be judged.

second, you got skipped verdict on first

Yeah yeah contest was slightly unbalanced but I still think it was a good contest, especially if it's his first one. Tbh we've seen much worse here and I actually hope that den2204 will be the problemsetter for some later contests too. I mean, problemsetters also have to learn how to do it properly and I believe this contest was actually pretty good for his first one. Gratz man !

I'm so regretful to spend 14 minutes to think for a solution in C.

Worse. I spend around 15 min think that error between 1.0000000 and 0.9999999 is greater than 1e-6. I have to actually put these values in formula to check the error.

You don't have to set constant value PI that much long

Well, I took it from my partner's template. Still I have a few things in my life that requires such a number.

hmmmm...

If you don't know that already you can also use

That's amazing!!!

Better than spending 19 minutes using tan instead of sin and don't know what's wrong :\

I somehow made the same mistake lol

which ide u used

I work with Kate text editor and Terminal.

Solution for D (if anyone is interested)

Move the king to cell (500, 500) in less than 1000 moves. Now board is divided into 4 squares by king's column and row. At least one of these squares contains no more than 166 cells. Then move king to the corner opposite to that square by diagonal. We will get to the corner in 499 moves, but the other player will have to move all >=(666-166)=500 rooks to avoid checks. So he will not have enough moves to avoid checks.

thought I was close to getting it during the contest but your solution seems much simpler than mine :()

Oh... you can move diagonally...

I don't like these type of tasks, where every number needs to be so precise.

In fact, it is possible to solve this problem in 998 steps with efficient implementation using the same logic, hence the number of moves is not

as preciseas you think.Moreover, 166 is not just an arbitrary number, it is rather an integer part of 666/4. Therefore you could have write 666/4 everywhere instead of 166 and get the same result without having to care about all the

precisenumbers.If

n=num_rooksandm=board_size(mis odd), then the condition for this to be a winning strategy was thatIn the current example, it boiled down to $500 > 499$, which is just enough for this equality to hold.

This is what he means by too precise.But almost all pigeonhole problems seem to have this property of just managing to work. So, I really enjoying this problem despite not being able to solve this.

Open Question: Is there any winning strategy if this inequality doesn't hold?Maybe you could generalize it, by checking the "quadrants" of every single position, not just the center. And then choose the closest one, instead of the center.

To me this looks like it's optimal, but still doesn't always win.

vjudge88's solution for problem F is a naive copy of https://blog.csdn.net/ShadyPi/article/details/79939990

Is there no hacking for this contest?I want to hack!!How much time after the contest can we submit solutions again?

After system testing completed

Anybody came up with a good solution for E ?

Binary search the answer, then check if there exist a cycle with all costs > mid?

Let's solve it with a binary search. We want to check that the answer is at least

k. Let's remove all edges with weight less thank(we can choose any direction on this edges). Notice that if our graph contains any cycles then the answer is more than k, otherwise the answer isn't greater thank.and how restore answer? if we change all edges with cost lower or equal to k its not create another cycle?

We know that there are no cycles, so we can find a topological ordering. After that we can direct our edges, which were removed, the right way

Can you please explain the use of topological ordering in more detail.

If we rearrange our vertexes in topological order then every edge will be directed from left to right. Some edges before removal were directed from left to right, we won't change their direction, but if an edge was directed from right to left we'll change its direction.

Miraak but why it will work .any proof for correctness.

and at what thing u r doing binary search. can topolozical sorting be applied on cyclic graph

Honestly, task C is nothing to do with coding competition. Very hard to understand problem statements.

Devic, Can I know why?

Yes, you can.

Oh, thank you

Not only because it can be googled. It's a pure geometric task which takes one line to solve in terms of code. There is no place for optimization, algorithmization, choosing the right data structure and so on.

I thought the same thing.

Basically it is more like D, E, F is too hard for most of Div.2 coders (including me. I ran into E for 80 min and I couldn't solve it). For them(from submission record, I think this includes both you and me), only meaningful problem was A, B, and C and one of them (C) were something like an basic geometric question for mathematics textbook or so. I agree with you — many people could write less than 5 lines to get it write after working on paper, especially when you write in python and don't really care about real number precision stuff. I think it would have been nicer if D was easier, so that more people could actually enjoy programming more.

It was impossible to hack! pretests are stronger than they suppose to be!

system testing started yes!!!

Yes!!!

Problem D can be solved without being able to move the king diagonally.

Works up until test #6 :)

Sorry this is not quite right. You have to move diagonally to the corner from the center. If you can't move diagonally, then you take one step either vertically or horizontally and terminate.

I got TLE on problem B :https://codeforces.com/contest/1100/submission/48351000

How do I speed this up?

Sorry, you just have wrong approach... Your works in O(nm) only

Thanks. I will try to get a better approach. Also, can you point out which part of the approach is wrong?

You seem checking every time if he can make contest. This is waste of information. Think about the case, when he knows that he needs 1, 2, 3, 4, 5 and he got 4. He don't have to re-check everything again. Right after seeing 4, it is already known that 1, 2, 3, 5 is missing without checking again.

Hi. I have modified the solution a little bit : https://codeforces.com/contest/1100/submission/48360651

Now getting a Runtime error. How do I test this?

Not quite sure because I'm not currently able to run or read carefully on your solution. Though I suppose the problem is that your vector val and group is both size m+1. Shouldn't one of them be size of n+1? Seems index out of range is being accessed.

Goddammit! Thanks. Got it. I guess this is why people allocate an array of size 10e5 and keep iterating over the fixed range. Miles to go.

:)

Glad you like it :)

D is a very nice problem. Thanks.

Pretest for E wast too weak

The worst contest in my life

For problem D got WA on test 40. Is there any way to know what that test was? Here is my submission.

I have the same problem as you. Has there been some modifications on codeforces which don't let you see the test you got wrong answer on ?

One difficulty is that this problem is interactive, so the input kind of depends on what you program prints. So you would need initial positions + the algorithm that moves the rooks.

Never mind, I found my bug: I have

but it should be

Argh!

Even the system tests are weak for E. (Attaching previous comment below)

Pretests of E are weak. I did not get TLE for the following DFS routine:

Good contest , just +1 shift was needed i.e D = E , E = F, questions were very good ! den2204

Now i have to wait 9 days for comeback (-_-)

My friend GreenGrape should teach den2204 how to prepare the pretests. Dear den2204, can you imagine that someone can use map to store the edges and write in the statement, that there are can be absolutely same edges, or at least include it in the pretests.

I agree, my solution failed exactly because of this issue, which is a shame because the problem was pretty nice. :/

Excuse me, I used C++17 to submit my solution during the competition, but I got Wrong answer. After I tried to use C++11 and C++14, I got Runtime error, so I want to ask this. Why? (My solution involves dividing by zero error)

C++11 Runtime error on test 6: 48359273

C++14 Runtime error on test 6: 48359354

C++17 Wrong answer on test 6: 48360585

Division by zero is undefined behavior in C++

That is to say, the error of dividing by zero in Online Judge may get uncertain results. Is that true? I always thought that if the pointer is out of bounds or divided by zero, it will return Runtime error, thank you.

Indeed, undefined behavior means at that point the program could do anything completely unexpected (i.e., "undefined"), and it usually varies depending on the computer and the compiler it's being ran on. So it doesn't make much sense to analyze WA, RE, TLE or whatever veredict whenever you're doing undefined operations. Another common one: accessing an array over its defined bounds.

Sometimes it might be Runtime Error because that sector of memory doesn't belong to you, sometimes it actually does and it's initialized with zeros, some others it belongs to you as well but it hasn't been initialized at all (contains a random value). All of these three cases could happen running exactly the same code.

Thank you for your help. I'll be more careful in programming and make fewer similar mistakes in the future. Another point is that I may not choose to use C++17 to submit the solution in the future. QAQ

How to solve E if to reverse each edge different controllers are used i.e. minimize the sum of cost all edges we have to reverse?

If we can solve this then same procedure can be used to solve it for directed graph with weights 1. The latter problem is minimum feedback arc set problem. This problem is known to be NP-hard.

I misread the problem during the contest but I think the problem might be reduced to minimum cost Hamilton path and thus NP complete(I don’t have a proof though).

Same thing happened with me also. I thought different controllers are required for different edges. :(

Here's my idea with proof for D: Greedy: Take the king to the centre. Then consider the number of rooks in each of the 4 squares formed by lines passing through the king. Look at the square with least amount of rooks. Walk the opposite way.

Proof: From the center, you can reach each corner in less than 500 moves. The square with least rooks must have number of rooks <= 666/4 by pigeon hole principle. So, there are atleast 500 rooks in the other 3 squares, combined. Now, Dasha has to move >= 500 rooks within < 500 moves, which is not possible. Hence, we are done.

Ah, so that's why those numbers were chosen. Very nice.

DELETEDI have a general question that is kind of related to this contest.

In today's contest there was a solution in my room that solved Problem B using brute force solution with complexity

O(n^{2}). The solution was supposed to give TLE on a test that has (n=m= 10^{5}) and all the numbers appearing starting from 1 and ending with 10^{5}.I wasn't able to hack this solution due to test case size which was about 512 KB, while codeforces seems to have the upper limit for a test case on hacking set to about 265 KB. Is there a reason for this low limit? Shouldn't it be a little higher? Specially since lately most problems seem to have bounds about 2.10

^{5}rather than 10^{5}like before, wouldn't this limit make it harder to hack solutions that fail due to TLE on large test cases?MikeMirzayanov, perhaps you could kindly provide an answer?

For hacks with test case size greater than the limit you can send the generator for your hack.

You mean sending the generator code rather than the test case itself?

Yeah, you can send the generator code instead

Aha! I asked a similar question to the authors during the contest and they replied "No Comments" to it. :(

I noticed that for the solution I submitted, some people who took longer than me scored higher, and some people who took half the time to answer that I did scored the same as me. How are points allocated for each question? I saw here that for problem A you lose 2 points per minute, but that doesn't seem to be happening for this contest.

For each failed attempt you lose 50 points

Got it. Thanks.

It's a shame C and E were well known. I liked those problems. (and I didn't know them, so I couldn't figure out E in time 0w0)

But nevertheless I think it is a good contest! Good job den2204

E well known.where? how to solve e

Binary search on the operators and then do this

but the above link describes the way in which we have to add edges , how can it be applied here

Imagine you want to check whether you can make the graph with C amount of operators.

Well notice that every edge > C is fixed and every edge <= C can be adjusted to fit the resulting graph.

If you can make a graph of edges with edge.c > C and find a topological sorting of the graph, then C works because you can adjust the other edges to fit the graph.

ok , so suppose we choose the best k .

then how to proceed further using topolozical sorting , how to revert edges so that graph becomes acyclic , since no. of reversions need not be minimum , so we can choose any edge .

but how ?

let us say we have the topological sort (I would suggest bfs based): {a,b,c,d,e}

And then we have this edge (b,d). We would not reverse that.

But we have another edge (c,a). It violates our toposort, so we would reverse that.

As for implementation you can store the positions of every element in the topological sort.

yes, as mentioned in G4G , we can assign undirected direction to all the edges with weights <= k , now we can check how to assign direction using top. order . this way graph remains acyclic

Problem D. Correct submission: 48360170 My contest submission: 48346037 This 1 character cost me 150 rating points (-65 instead of about +90). The worst part in it, that it passed pretests when it is wrong 1 out of 4 times. This should have failed on pretests. But now I am in the same position as people who couldn't solve D. I know that I made a terrible mistake with that x instead of y, but I still have the question: Why does Codfeforces judging system is so unfair? EVERYONE who gets wrong answer on pretest have the chance to correct it, while EVERYONE who passed pretests with a wrong submission will lose ALL of the points for that problem. This is just a game of luck. In My opinion there would be 2 fair way to judge with a similar system. A: There are no pretests, only maintests. That would be much harder to make a good solution, but it would be fair. B: pretests=maintests. This would make a little easier to make a good solution, and it would be fair as well.

My wrong submission for D that I skipped during the competition passed system test. Paradoxically I would be first if I didn't come up with correct solution.

Wrong submission: https://codeforces.com/contest/1100/submission/48362464

I can only handle A,B,C after contest.Bt during contest time I was a great looserThere is a problem that harder than F and I revise it wrong :( Actually if I make the array in size 1e6 then I can get Happy New Year :(

I wanna ask sth about E, i use the dfn[], get the time when i into a node, but others get the time they leave the node, as a result i get WA, i hope someone can explain it. Thank you so much.

i use dfs to solve,bfs get in or out can solve it,so i wanna konw why use dfs to get the into time cannot solve it

For problem F, are we supposed to consider the binary representations of each number? I really want to learn about that.

Please add the editorials as soon as possible.

I know it's an accident but problem F really coincides with a problem in a training contest proposed by whzzt called "异或"(XOR, in English).

In that problem, you will be given a sequence of n integers a[] and m queries {li, ri, di}. For each query, you need to choose a subset of a[li, ri], compute the XOR-sum of all numbers in this subset and XOR the result by di so that the final result is maximized. You need to output the final result. Constraints: n, m <= 10^6 and 0 <= ai, di < 2^30.

Anyway, D is awesome.