Hi!

*Summer Informatics School (SIS / LKSH) is a summer school for students in grades 6-10. SIS is focused mainly (but not only) on students participating in the Olympiads in Informatics — from beginners to participants of international competitions. SIS is held in July and August, each of them is visited by about 200 students from all over Russia and abroad. The language of the camp is Russian. Additional information about SIS is posted at lksh.ru*

Right now, the August branch of the Summer Computer School is running, and on August 11 the traditional team Olympiad is going to place. I am happy to present a rated round based on it!

The round will be rated for both divisions, will be held in Aug/11/2018 16:35 (Moscow time), in each division there will be 5 tasks and 2 hours to solve them.

The problems of this round and the SIS olympiad were authored and prepared by SIS's teachers: izban, achulkov2, Schemtschik, peltorator, craborac, asokol, Morokei, Burunduk2. Also, I would like to thank Kurpilyansky for his help with olympiad's organization.

Thanks to our problems testers: meshanya, Burunduk2, vintage_Vlad_Makeev, niyaznigmatul, manoprenko!

Thank you, MikeMirzayanov, for the codeforces and polygon systems!

Yes, **we are aware**, that this contest clashes with ProCon Junior on codechef. However, given the schedule of the SIS and the approaching VK cup finals we can't do anything with it, sorry for that =/

Good luck!

**UPD**: The round will have one interactive problem for both divisions. Please, read the post about interactive problems here: Interactive Problems: Guide for Participants.

**Congratulations to winners!**

Div1:

Div2:

**Upd** Thank you for participation! Due to vk-cup conduction rating recalculation will happen slightly later, than usually.

**Upd** The editorial is published!

You may also check the unofficial editorial, written by neal.

Ok so now CF will host contests prepared for middle school students. Kill me now!!!!!!

Don't underestimate the complexity of this contest :).

I hadn't been surprised if you would solve nothing from a SIS hard olympiad :)

Says the guy who has solved only two problems in CF and has rating <1500 (No disrespect to lower rated guys but being that arrogant calls for such a response)

Why underestimate middle school students? There are quite a lot middle school students who rates more than 2400 on Codeforces.

This will hopefully be a good contest for those of us training for olympiads :)

Which problems are going to be based on the olympiad?

Gotem??

hi sguu

## But Is It Contributed?

## Or Is It Rated?

The start time changed from 16:05 UTC+8 to 21:35 UTC+8... At least I can go to bed before the next day.

I thought that I can participate in this Round before I noticed that the start time had changed.

The disadvantage of training during summer vacation in boarding schools.

mathforces?

round #503 and error #503(unavailable Gateway).. another coincidence??... hopefully..not this time :)

Shouldn't it be 502 Bad Gateway & 504 Gateway Timeout?

(UPD: Comment of arbitrary_A is fixed)

<del>(BTW, why didn't the server down in Round #500(Server Internal Error)?)</del>

hope #service unavailable fact... btw error may b unpredictable...

I saw 500 error yesterday, it lasted for a very short time, then changed into 502 error, and then the announcement.

I just said the truth I saw that had already happened. I also hope that it will not happen again.

I guess despite the original contest being IOI styled, this round will be CF styled, as always?

Original contest is ACM styled like most team olympiads.

Hope to be a good contest as it seems to be, wish you all guys better ratings...

You don't know how much I love you, when you down-vote for nothing!!!

I wouldn't recommend you posting here without any serious reasons. Any controversial and/or "meaningless" comments are very likely to be downvoted.

I don't know how those memes can be serious, anyway I won't comment anymore...

Great to see izban is one of the problemsetters!

Are there any shared problems between both divisions ?

As always

Please provide some more tutorial about an interactive problem.

Please...

http://mirror.codeforces.com/blog/entry/45307

Everybody jump on the downvote train!

Great round! The time really fits me.

As the prizes of the previous round is postponed, will there be prizes in this round?

I think it will be on div1 + div2 round so no.

,

I hope it won't be unrated like Round #502 (Laugh

Oh ,hope that won't happen again.

this blog full of downvote comments

Why are you people downvoting every single comment? Just for fun? It completely ruins the whole discussion, because if anyone says anything, he will get downvoted to hell.

Is it disrated on downforces?

Why most of the comments in this post are hidden?I can't find the reason why you downvote so much.(Because of this I upvoted all the comments)

And I downvoted your comment

Why are all comments downvoted?

Someone made a bunch of alts to downvote, probably. If not manually, I'm guessing someone used a bot. (I'm surprised codeforces doesn't have some countermeasure this)

Codeforces uses reCAPTCHA to prevent bots, but it can be passed automatically with little money.

Hosting VK Finals now, I do not have enough time to investigate the situation. But definitely they are not simple bots. Probably, it is some kind of flash mob or hacked accounts.

So how to deal with these downvotes?

Keep them polluting the contrib. of everyone?

please delay/starts it 15 min later than scheduled time because there is a contest at atcoder also.. please this is a request.

you can AK ABC in 85 minutes and take part in CF

...

Hi

I made 10 accounts in Codeforces and downvoted some comments here, and of course others saw that and kept downvoting the comments. Some of you may already knew this but my purpose was to show that lots of people here are just dumb followers and they don't have any brain to decide what to do, they just copy others' actions.

So to the guys who got downvoted in this community or as some guy in other blog said

are not a clownto make others laugh, don't worry, it's not your issue.Thank you for being idiots.

Did you know that you are so funny?? That's not being an idiot, when someone sees everyone is downvoting him/her, he/she starts downvoting others... you are the only idiot here... Hey idiot, now go and make more accounts and downvote me..

They did not claim to be funny.

Why would that make sense?

Mob mentality/conformity, people follow the crowd, when they see a lot of downvotes they are already ready to agree with the majority. I think this is a real effect, but I don't think that downvoting everyone to "prove a point" causes any real change... it is just an annoyance

That is not what I asked. Tour_guide claimed that when someone is downvoted, they start downvoting others.

Sorry, I misunderstood. I suppose he means that person gets very mad and wants to take it out on others, but I think most of the codeforces community is better than that

Why not?? Don't you feel bad when everyone is downvoting your every single comment?? Less people can accept others disagreement and don't react...

I might, I just don't see how downvoting others would help whatsoever...

I don't really know and can't explain, something instinct, I think. maybe you're right...

So do you think this is the right way to tell them they're wrong?

Good job. Most posts above had 50+ downvotes even though you casted only 10. Time for people to wake up.

Why someone was so bored to downvote everyone?

I think that means bored.:（

I have short by the way.

That one person who'd like to comment but too afraid of the downvotes.

It's lucky that you still have the chance to be afraid.

If you want to see the comments in this thread without clicking on every link or you have a strong will and always want to see every downvoted comment, then install my brand new super duper userscript created in the last 5 minutes: https://openuserjs.org/scripts/mraron/Codeforces_Show_Bad_Comments (you may need something like tampermonkey addon to get it working though)

[/END of self advertisement]

ads are getting smarter

Hope the description of the problems can be concise.

Scoring?

Will be announced later.

Anyone join the campaign to turn all comments to upvote (except those ask for downvotes or unhelpful comments) ?

Freaking A >(

problem c is much harder than problem b

How to solve problem D?

You can calculate the cost for other have less or equal than k votes, and i have greater than k vote in

O(NlgN). then test for all k = 1 to N/2+1----------------- Sorry I confused C/D

If N % 4 != 0, answer is -1

If N % 4, Let's think about the sequence of A[1 to N/2], A[N/2 to 1]. ex) 1 2 3 2 3 4 3 2 1 -> 1 2 3 2 3 / 3 4 3 2 1

Then Since diff is 1 or -1, it it guaranteed existance of answer. Then you can find it by binary search. I want to explain more but i should leave now sorry :(

How to solve div.2C?

Iterate the number of votes we want from party 1, then get the minimum cost required for this number of votes to be the majority. Answer is the minimum of the result from all the different number of votes required.

1 <= k <= n, you need to calculate cost (k = 1; k <= n), k is votes, then min of these cost is our answer, firstly you should cut k <= votes[i] (1 <= i <= m) x = votes[i] — k + 1, then cur_votes += x; second need_votes = k — cur_votes then first sum of need_votes added your cur_cost

Interesting problems, how to do Div1C?

For div2c the idea is to always select the minimum voting cost until it's time to choose the last voter... is this approach right? and if it is, how to implement the last part of it ?

Div.2 C is bunch of priority queues and sets and maps. I was not able to finish it in time :'(

You can do it without any of those things.

I did not come up with a more simple idea so I wrote a bunch of these things mixed. XD

Does every interactive problem on CF have to be binsearch :P? (OK, I know there are many counterexamples, but number of such problems is too high and they are rarely interesting)

This was exactly what I thought when I saw 60 in problem B :)

I didn't know how to solve. Saw 60, and that two queries are definite for as for i and i+n/2, I was sure it was Bsearch. I coded for both way, one was working, submitted and accepted. I don't even how that solution is correct but its working.

Any idea for Div1 C?

Btw, C is a bit too hard, but today's problemset is good. Problems are balanced between thinking and implementing. I enjoyed B a lot.

`Btw, C is a bit too hard, but today's problemset is good.`

The kinda converse to Div 1 C (that too not constructive, but just the existence of such sets) came in Tuymaada 2010 Junior Problem 8 but it's still unsolved on AoPS !

I hope the answer to this problem (Div 1 C/Div 2 E) will give some hint on that problem, because that one is bugging me for a while.

This seems to be exactly a constructive version of Chvatal-Lovasz theorem about semi-kernels in directed graphs.

The simplest proof is indeed constructive and should yield a linear algorithm: pick some vertex

v, delete it and its "positive neighbourhood" (the set of all verticesusuch that there exists av→uedge), recursively find a solution in the remaining graph, addvback if you may do that without breaking independence of the set of the picked vertices. Now you have a solution for the original graph (to verify that, just consider two cases: you addedvback or you did not).Link to the original paper: Chvatal-Lovasz theorem.

For implementation details, you may refer to this submission: 41500579.

can anyone tell me an idea/solution of pb "C. Elections" in recursive + memorizing

You mean DP? Because I'm pretty sure a DP solution is not available for Div 2. C, as in the states you don't know which party have how many votes in order to reach for an optimal state.

My story in A

Unlucky round for me.

I started by quickly solving A, then moved to C. Wrote a solution in 20 minutes, and only then red the second sample. I really need a browser extension to replace "loop" -> "self-loop". Then I moved to b, and got 2 WA's in test 16. What is the problem?

I solved the problem zero-indexed, so I had to fix the indexes by adding 1. But if i = 0, it outputs 0 giving WA. I didn't realize this was the mistake, shuffled some code around, and got "pretests passed". Then only later I realize this and had to resubmit.

But yeah, I couldn't solve C, so I don't mind losing rating.

Does anyone know what's the issue causing WA on pretest 5 in div2C?

You get WA with below input.

2 2 1 1 2 10

for(int i=n-1;i>=0;i--){

*for(int i=n;i>=0;i--){

Maybe you select the voters who originally support Party 1 to let them change. Before I fix it, I got a WA5

Hi i thought I found someone violating the rule of codeforces contest

higu and adityasr share the same solution for Problem C of Div 2

This the first time for me to enter a contest , in fact I couldn't solve any problem as I misunderstood many points and also I haven't practice a lot ,but I hope to do better than this the next time.....

See!!

I told you everybody that Benq is going to be the next world conqueror of CP.

rip everybody

Benq wil beat you all!

This is false; C was incorrect. :/

your alts are all boring srsly

could any one tell me what is that system testing that have started after finishing the contest,please??

In the pretest, only part of test cases are run. After the contest, the system judges all the solution with all test cases and test cases used in hacking

Thanks a lot! ,I do really appreciate your comment....

I think today's problems are very nice.

But,I think the difficulty of solving B and the difficulty of solving C is too different.QAQ

How to solve Div1 C? I passed pretest but I'm not sure my solution...

Is DIV1A/DIV2C same as http://mirror.codeforces.com/contest/458/problem/C

That's crazy. Even the same legend. After more than 1000 contests, it must be really hard to find these coincidences.

I think every time there is an election in Russia, someone "comes up" with that problem. That's why it has already been in a contest.

lol. Cool satire.

The main difference is the constraints. My O(n^2) solution was accepted, but in this problem, it will fail as 1<=n<=10^5.

We were not aware about this problem, sorry =/.

For div2 C, first lower down every other party to at most k votes, then we can get enough votes from those who need least without considering if he is from the most voted party. Check every k from 0 to m, we can find the minimum cost.

I pass pretest with this solution in 1900ms, and reducing time to 1100ms using priority_queue.

Hope it won't get a TLE on the system test :p.

Wrong answer on test 21……QAQ

I should check every k from 0 to n, not m……:(

Hi. What is K? and why do we have to reduce other parties? Can you please elaborate?

The number of votes that you give to party 1. If you fix the number of votes, it is easy to calculate the min cost. So fix the number of votes (k), but try all possible values.

Did someone passed Div1D with

O(n^{3})?Just found 41488961.

I did, but it took quite some iterations (== random changes to the code) to make GCC on codeforces to compile it as I want...

It would be awesome if the "run on server" tab provides an option to download the compiled binary. One can at least see what exactly the compiler is doing.

Yeah, but in my case, with a dirty trick. See my comment below.

Yeah that's insane. I have to trust more in CF servers.

As ilyakor's comment above tells us, trusting can be pretty dangerous. It's best to be sure, and testing in a local environment similar to CF servers helps greatly.

In div. 1 C, about 70% submissions that passed the pretests, failed later on test 20. Why wasn't this test added to pretests?

To don’t help people with heuristics I guess. Pretests should help you catch bugs.

Maybe I don't understand the idea of pretests, but in my opinion, they should help contestants check if their solution is mostly correct.

In general, I believe that most heuristics and/or brute-force solutions should fail on good pretests. Also, stupid corner cases should be included in the pretests as well.

No, pretests should give you some idea if a solution is correct. You can catch bugs by stresstesting, but can't realise you're missing something essential by doing that.

Only if you expect a lot of solutions with heuristics that wouldn't pass some type of tests, which isn't very realistic in problems like this one. I'm sure there weren't many heuristics among the failing submissions.

Looks like most of the solutions to Div1 D (1019D - Большой треугольник) are with binary search. I didn't come up with the right solution, but managed to squeeze an

O(n^{3}) one (41491248) instead, using 1950 milliseconds out of 3000. It involved a trick and a dirty trick. I'd like to elaborate on that, to increase awareness of such tricks for contestants and problemsetters alike.The

trickis just loop unrolling, which is around this part:The

`static foreach`

is expanded at compile time into 4 blocks with r = 0, 1, 2, 3. In C++, one way to do it is to use a`#define`

inside the loop, and instantiate it manually 4 times instead.The

dirty trickis doing all of the above in signed`int32`

: as the program is run in a 32-bit environment, it helps greatly. If we find the triangles which have area equal tosmodulo 2^{32}, only then we check again using true but slow`int64`

multiplication and comparison.Now, as a problemsetter, you can't really do anything against loop unrolling. In fact, in simpler cases, a good optimizing compiler does that itself under the hood.

Against the using

`int32`

part though, it is possible to construct a test case. Make all coordinates divisible by 2^{16}, andsdivisible by 2^{32}. This way, all multiplications modulo 2^{32}will result in zeroes, and the actualO(n^{3}) work will be done by the`int64`

part of the solution. In this case, making it run for five or six seconds.I also did it in O(n^3), and without the dirty trick (I did consider it, but figured it was at risk of being attacked). My tricks were to - use pragmas to enable more aggressive compilation - precompute the signed area of triangle (P, Q, origin) for every P and Q, after which the area for a triangle (P, Q, R) can be computed purely with addition (which is less expensive than multiplication on a 32-bit system).

It would be nice if Codeforces would move to 64-bit one day — these days even the chips in your mobile phone are 64-bit!

Argh. I wrote essentially the same code as you, but without pragmas, so it got TLE. I guess I will know what pragmas to use next time.

(I also tried putting the points into buckets depending on the coordinates modulo 11, and then only comparing those buckets for which the area is correct modulo 11. But it got WA due to a silly bug.)

mind sharing any good resource on learning these pragmas? Thanks!

I read them in a comment on some previous contest (might have been a different architecture pragma, I couldn't remember so picked something that looked sensible), but I don't recall which contest it was.

Pragmas are instructions or information for the compiler embedded in code, e.g.

`pragma once`

(ignore all`include`

-s for this file except the first one) or`pragma GCC target()`

for setting target architecture. Obviously, they're compiler-specific, so you have to read the documentation for a specific compiler. For example, I found this in about 5 seconds, so don't ask for shit you can look up yourself.Function attributes are more interesting. You have the optimize attribute, with which you can tell the compiler to aggressively optimize just one function — very useful, since

`Ofast`

on everything can slow down some other part of your code. Then there's always_inline attribute, with which you can force inlining of a function that the compiler would usually skip because it's too long (e.g. a lengthy arithmetic expression that's used very often).Thank you for the useful information. Sorry that you get downvoted for helping :(

I did the same with you, but got TLE with C++11. After the contest, I submit the same code with C++14/C++17 and get a Accepted. :(

What do you think about the following idea: pick random p < 2^16, begin by taking coordinates mod p, and compute hash of triangle areas mod p using integer type.

EDIT: Never mind, mod is probably slower

EDIT: okay, seems like Gassa got it to work.

I tried it now but realized that, yeah, we seem to need a modulo operation in the innermost loop.

I've tried to make the calculations modulo

`mod`

-squared. But, if we take the coordinates modulo`mod`

, their product is correct modulo`mod`

, but not correct modulo`mod`

-squared. For example, 1 (modulo 2) times 1 (modulo 2) can in reality be either 1 or 3 (modulo 4), we don't know. So, seems we have to use the slow`%`

operation.What about a memory access, keep array f[x] = 1 if x is a multiple of p and 0 otherwise?

Now I'm confused. How would it help?

The example modulo 4 does not work even for non-multiples of 2.

No, just do the original thing, to check if area + S is a multiple of p, just access f[area-s]. (maybe pick p < 2^12 and have bitset of size 2*p^2?)

Hmm, that could have worked, but doesn't for me (attempt: 41503640).

When the

`mod`

is too small, we would have too many false positives. When the`mod`

-squared is too large, the array does not fit into processor cache. I've got TL with`mod = 997`

.Actually, scratch that. With a bit of refactoring and

`mod = 107`

, this passes: 41503762. Thanks for the idea!This can still be easily hacked. (it's mostly safe against system test, though)

Maybe we could do something like this:

Pick a

randomp ~ 2000. (see note)As suggested, we want to find all solutions mod p (of which there are hopefully ^3/p) and test them individually.

Our basic loop is as follows:

for each pair of points (P1, P2): determine all P3 which gives right area mod p test these triangles

Let ABCDEF be some constants. Let P3 = (x, y)

Now, after fixing P1 and P2, we know that the area mod P is given by Ax+By+C. We want Ax+By+C = S. If A=0, we handle it separately. Else, we divide the equation by A, giving x+Dy+E = S, or x+Dy = F.

To find all such points such that x+Dy = F, we can precompute x+Dy for each point and each D. That is, we can store a vector of points for each pair (D, F), representing points such that x+Dy = F.

The complexity for random points should be O(n^3/p + np). Choosing p~n gives O(n^2).

Note: Of course, a given p can be easily hacked. Just choose S = p^2 and all coordinates multiples of p.

I'm not sure if a randomly generated p can be hacked; there are not that many good choices of p too.

I got TLE with O(n^2 log n) in Java during contest so was thinking the TLE was too tight, but I guess this proves me wrong.

Are you planning to change this feature/bug with bomb on sample other than the first one? It cost me a victory...

It's truly a feature. It's meant to prevent you from failing in very simple ways as sending wrong file or wrong formatting of output/input. It's not meant to not penalize you for WA on third sample.

I was formating output incorrectly on second sample.

Obviously wrong formatting of I/O. Or submitting a wrong solution or forgetting to delete random input generator from your code or other silly mistakes. It's extra lenient, but worth having despite that IMO.

How to solve Div1D

Just brute force, as discussed above.

Nice TOP3! High five Marcin_smu and Radewoosh :))

Btw, it seems that many people tried some heuristics to C and all fo them failed. I have to admit that I also pushed some shitty solution... But it seems it was less shitty than others :P. I basically keep all the time some set of vertices that is an independent set and when I see a vertex that is not reachable in two steps from others I delete all its out-neighbours from set and add itself. And I do it until I run out of time or until I get a solution — whichever is sooner. And it luckily turns out that second case is always sooner :P

Don't piss me off...

Fortune favors the bold xD

Nice! I admit I was lucky!

Why doesn't a problem like this with lots of heuristic solutions just have all directed graphs of size <= 5 as tests? There are only (2

^{4})^{5}= 1048576 directed graphs of size 5, so filling every test with as many size-5-or-less graphs as possible would only require like 10 tests.I opened problem E after the end of the contest, and found it was actually an extension of my recent CSA problem. (That extension completely changes things, so it's not a copy) Although I don't have any good ideas, I regret myself of not opening E :)

And big thanks to setters for the great round! Problems were really interesting. It looks even better because last round was very poor. Thank you so much for the effort!

I'm currently suffering from Post-Codeforces-Round-Depression :P

I'm suffering from Continually-Refreshing-The-Page-Every-37-Seconds-Even-Though-I-Already-Know-How-Badly-I-Did Depression :/

Im suffering heavily from post system test depression....

Runtime error on codeforces, same test case works perfectly on my pc. Same compiler. This isnt first time its happening. just wow

That happens, for example you over bounded when seeking an element of an array and there happens to be a zero (or anything "make sense") on your PC.

The compiler option may be different. Try to compile it with "-O2" ?

Run your program with Valgrind with a small input, you'll probably find you're reading something out of bounds or uninitialized. Usually that happens to be a 0 in your PC but random data in CF, so it seems to work but doesn't

I solved it, thanks everyone :)

Rating changings ?

For anyone interested, I wrote up solution sketches for a few of the Div 1 problems from today -- take a look: http://mirror.codeforces.com/blog/entry/61144

Editorial ??

Can anybody tell me what is the complexity of this code? for problem

Div2 CMy idea : To get answer there are only 2 possibilities

1) We change the perception of person with minimum cost

2) We change the perception of

person with minimum costof the party which has maximum followers.I am using string

`s`

to store which people are followers of our party(i.e. party 0).Then either the first condition holds or second conditions holds for getting the correct answer.

I think to find the answer we need to change at most 1500 characters in s and so my solution should work in

`O(2*1500*N)`

where N = overhead for calculation of each dp =`O(n)`

. So final solution should work in O(n^2).I don't know why it is showing

TLEontestcase 46which then I had to add an exception in code:p and get it accepted.I'm finally back to red and not at all happy about it. I'd much rather get a good place because I did well than because almost nobody did well. What's even worse, D is solvable by very optimised bruteforce, A is a ripoff and C, D are "write a heuristic, maybe you'll pass lol" problems.

Well, unfortunately we underestimated the squeezability of the dumb solution in D.

But what is wrong with C?

http://mirror.codeforces.com/blog/entry/61087?#comment-450850

Also the number of failed solutions on tests 20-21.

Got TLE for three times on div1.A with an O(n^2logn) solution, which made me upset and unwilling to do anything...

Is the tutorial of the round coming soon? It seemed that some solutions of Div2.D can be hacked.

For example, use this data to hack wnsh's solution:

or use this to hack usachevd0's solution:

So I have a little doubt that if the data need to strengthen.

upd: I am so sorry that I had not seen the constraint condition that |a1 — an| = 1.difference between first and last must be +/- 1 (it's a circle)

Yeah >u<

Please, share the editorial of Round #503(SIS Div.2)

Why this submission works weird? I submitted the program that only prints

`? n`

6 times, and then`! n`

. But the judgement protocol shows me that for the first testcase the program(submitted) prints 7 and 6. There isn't even`?`

character.For the second testcase, the program is expected not to print

`-1`

at all, but the judge says so.Did I misread the problem statements?

The first value is the answer. The second value is the number of times you asked.

You mean,

`7 6`

from "output" section of Test #1 means "the answer" and "# of queries"? But I can't understand why the value`7`

pops up. The code only prints n, which is 8 for the first test...Also, the second case gets me confused even more. The code never prints the answer, -1. But the judgement says it does, and the test is passed.

The first value is the answer(0-index). for example, with test case 1, first value must be 3 or 7.

about second example, you should read your code carefully.

OMG... Sorry for my carelessness.

The root of the problem was my mistake from code. I wrongly printed "the value" instead of "the position".

To my shame, I doubted the judging system. Thank you Oumae_Kumiko !!

Please, share the editorial of Round #503(SIS Div.2)

Thank you for preparing the problems and round. :)

Hello there, I'm new to Codeforces. Few days ago I tried to take part in this contest as my first one. When the contest is finished, I checked my code again but couldn't find any problem, so I asked my friend to help with me but we found that

with different compiler selected, the result is different.Why and is it unfair?The following pictures shows that the result of same file submitted with different compiler. My code could be viewed/download here

After the contest you can see the test case your submission failed on at the bottom: http://mirror.codeforces.com/contest/1020/submission/41532464

It looks like

`abs()`

is returning a floating-point value for you in the older version of C++, but it returns an integer in C++11. In particular, this line is the issue because you are actually printing a floating-point number:Also, I recommend against using

`endl`

. See this comment for an explanation.well,I'm still waiting for the solution for this contest,it's so slow QAQ

you can see them here https://mirror.codeforces.com/blog/entry/61161

wow!Thanks for your help!

Why so many downvotes here?

Maybe something wrong with the checker of problem D?

The coordinates are all integers, so the area of the triangle should be xxxx.0 or xxxx.5.

But why the checker says the area of my triangle equal to xxxx.1?

Submission here: 137387839

cdkrot