Hello Codeforces!

Grimoire of Code, the official Competitive Programming club of IIT Kharagpur, is happy to invite everyone to take part in Codeforces Round 819 (Div. 1 + Div. 2) and Grimoire of Code Annual Contest 2022 which will take place on Sep/06/2022 17:35 (Moscow time). **This round will be rated for everyone.**

You will be given **8 problems** and **2 hours 15 minutes** to solve them.

The problems have been authored and prepared by Grimoire of Code members Anubhav anubhavdhar Dhar, Debajyoti little_angel Dasgupta, and Mainak Newtech66 Roy.

We'd like to thank all the people who made this round possible:

- Artyom123 for their supercalifragilisticexpialidocious coordination + Russian translations!
- shivansh1102 for being the one and only leamder. No, "leamder" is not a typo
- harsh639 and rishit3234 for behind-the-scenes work and proposing tasks that did not make it to the final problemset.
- All of our testers: AwakeAnay, TimeWarp101, naman1601, preyam, fugazi, kasparovian, amul_agrawal, TheOneYouWant, AlbertoTDNeto, thenymphsofdelphi, physics0523, Um_nik, Absyarka, MayankAg, hackerbhaiya, Pratyush1606, sqrt_pi, for testing and providing useful feedback.
- MikeMirzayanov for great platforms Codeforces and Polygon.

We hope everyone will enjoy the contest.

See you on the leaderboard!

*About Grimoire of Code:*

Grimoire of Code is the official Competitive Programming club of IIT Kharagpur created by and for competitive programming enthusiasts. We promote competitive programming culture in our college, and provide a forum for interested minds to discuss their thoughts and ideas. We also conduct mock coding rounds for placements and internships in Indian colleges.

You can check out our Facebook page **here**.

You can check out the problems from last year's Annual Contest **here**.

## Updates

**UPD1:** The contest duration has been extended to **2 hours 15 minutes**.

**UPD2:** Score distribution: $$$500-1000-1500-2000-2250-2750-3250-3500$$$

**UPD3:** Congratulations to the winners!

**UPD4:** **Editorial** ~~ (editorial for F will be added soon) ~~ It is now added.

**UPD5:** The contest is unrated due to problem copying. Details.

As a testerwhen asking for downvotes and asking for upvotes are unsurprisingly in the same SCC

As a participant, I will have participated.

Spoiler for upvotes:Spoiler for upvotes:Spoiler for upvotes:Spoiler for upvotes:Spoiler for upvotes:Spoiler for upvotes:Spoiler for upvotes:Spoiler for upvotes:Spoiler for upvotes:Spoiler for upvotes:Spoiler for upvotes:Spoiler for upvotes:Spoiler for upvotes:Spoiler for upvotes:Spoiler for upvotes:Spoiler for upvotes:Spoiler for upvotes:Spoiler for upvotes:Spoiler for upvotes:Spoiler for upvotes:Spoiler for upvotes:Spoiler for upvotes:Spoiler for upvotes:Spoiler for upvotes:Spoiler for upvotes:Spoiler for upvotes:Spoiler for upvotes:Spoiler for upvotes:Spoiler for upvotes:Spoiler for upvotes:Spoiler for upvotes:Spoiler for upvotes:Spoiler for upvotes:Spoiler for upvotes:Spoiler for upvotes:Spoiler for upvotes:Spoiler for upvotes:Spoiler for upvotes:Spoiler for upvotes:Spoiler for upvotes:Spoiler for upvotes:Spoiler for upvotes:Spoiler for upvotes:Spoiler for upvotes:Spoiler for upvotes:Spoiler for upvotes:Spoiler for upvotes:Spoiler for upvotes:Spoiler for upvotes:Spoiler for upvotes:Spoiler for upvotes:Spoiler for upvotes:Spoiler for upvotes:Spoiler for upvotes:Spoiler for upvotes:Spoiler for upvotes:Spoiler for upvotes:Spoiler for upvotes:Spoiler for upvotes:Spoiler for upvotes:Spoiler for upvotes:Spoiler for upvotes:Spoiler for upvotes:Spoiler for upvotes:Spoiler for upvotes:Spoiler for upvotes:Spoiler for upvotes:Spoiler for upvotes:Spoiler for upvotes:Spoiler for upvotes:Spoiler for upvotes:Spoiler for upvotes:Spoiler for upvotes:Spoiler for upvotes:Spoiler for upvotes:Spoiler for upvotes:Spoiler for upvotes:Spoiler for upvotes:Spoiler for upvotes:Spoiler for upvotes:Spoiler for upvotes:Spoiler for upvotes:Spoiler for upvotes:Spoiler for upvotes:Spoiler for upvotes:Spoiler for upvotes:Spoiler for upvotes:Spoiler for upvotes:Spoiler for upvotes:Spoiler for upvotes:Spoiler for upvotes:Spoiler for upvotes:Spoiler for upvotes:Spoiler for upvotes:Spoiler for upvotes:Spoiler for upvotes:Spoiler for upvotes:Spoiler for upvotes:Spoiler for upvotes:Spoiler for upvotes:Spoiler for upvotes:Spoiler for upvotes:Spoiler for upvotes:Spoiler for upvotes:Spoiler for upvotes:Spoiler for upvotes:Spoiler for upvotes:Spoiler for upvotes:Spoiler for upvotes:Spoiler for upvotes:Spoiler for upvotes:Spoiler for upvotes:Spoiler for upvotes:Spoiler for upvotes:Spoiler for upvotes:Spoiler for upvotes:Spoiler for upvotes:Spoiler for upvotes:Spoiler for upvotes:Spoiler for upvotes:Spoiler for upvotes:Spoiler for upvotes:Spoiler for upvotes:Spoiler for upvotes:Spoiler for upvotes:Spoiler for upvotes:Spoiler for upvotes:Spoiler for upvotes:Spoiler for upvotes:Spoiler for upvotes:Spoiler for upvotes:Spoiler for upvotes:Spoiler for upvotes:Spoiler for upvotes:Spoiler for upvotes:Spoiler for upvotes:Spoiler for upvotes:Spoiler for upvotes:you've reposted in the wrong connected component

For not wasting time, the secret in lots of spoiler is help_me_please. :D.

posting mind blowing meme to get upvotes :

As the leamderGive me upvotes :)

DomneAs an authorHope that everyone will enjoy the problems! :D

Div 1+2 and 8 problems! Looking forward to it.

When will scoring distribution come out?

As a tester, I just hope everyone's super-super-supercalifragilisticexpialidocious performance on the contest.

As not a tester, I did not test this round.

I want contribution, I also want rating, I can't have both of them, So I choose rating instead of contribution.

As a participant,I'm hoping for a great round!

As an author, me too.

Looking forward to it :p

As an authoromg that thumbs up emoji looks fabulous

also YOU SHOULDNT BRAINSTORM THE SOLUTIONS WITH OTHERS DURING CONTEST ITS AGAINST THE RULES

(yes, I couldn't believe my eyes after reading that line)

what's wrong with all this downvote, I'm just stating the facts. are they all cheaters or what

Come on dude! Isn't it obvious that, by "you guys", I meant everyone, brainstorm the solutions "individually"

Well, yes, except for the fact that brainstorming, in common sense and as stated on this wikipedia page, is a "group creativity technique". Using a group creativity technique where there isn't a group would be a pain...

got downvoted for being right

now this is a comeback

help me, this comeback has reversed somehow

after the contest, I realized.

You (plural) might have tried your best to make the problems as interesting as possible, but you (singular) didn't! Shame on you!F*ck you problem stoler.

Let's downvote him to make his contribution lower than zxyoi cuz it's the second time such situation happens

IIT KGP !orz

What is IIT KGP?

Indian Institute of Technology, Kharagpur. It's India's one of the best engineering institute.

As a pupil contestanti hope not to lose points in this contest

Lol bro, u r on the edge of pupil-newbie

and what's so wrong with it?

i lost 70 points this round. now it's unrated i got my points back (Thanks god)

why is it unrated?

one of problems was copied from another judge.

I hope I can do at least the first one

When the guy who steals your meme gets more upvotes than you.

Reference: https://codeforces.com/blog/entry/102996?#comment-914421

first time ?

Really happy to see IIT KGP finally setting contests in Codeforces. Kudos to the entire team of problemsetters.

I love div.1 + div.2 contests

Good luck for everyone!!

Hoping for a nice problems❤️

Good luck

Is it rated for DIV 2 ?

Yes, it is rated for all divisions.

not any more xD

Hoping to solve 4 problems, would be great

Wishing you luck, to solve even more

Hope you can solve more problems too

Hope to become as soon as possible Candidate Master.

All the best

As a xxxGive me xxx

Here is: hided content

However, A new comment will be shown in blue background

hidden content*

good luck everyone and hope the problems this time is not mathforces qwq

some moderate amount of math is fine, last time it was kinda excessive but the problems were very interesting

god, you're everywhere

Contest time collides with IND VS AFG cricket match. Very unfortunate.

It's IND vs SL!

All the Best EveryOne!

Good luck to all

Sometimes you win, sometimes you learn

We just got trolled

How to solve D?

Color everything in blue. Now do a DFS from node 1. Since m is at most n+2, you will get at most 3 back edges. Store those back edges while traversing by dfs.

Now you have obtained some tree rooted in node 1, with some back edges. Now if you have less than 3 back edges, simply color those back edges as red.

If you have 3 back edges, and they form a triangle, in other words have form a-b, b-c, a-c, color a-b and b-c as red. Now look at a-c. Coloring this as red would be a waste, because a, b, c are already connected. So instead leave it as blue. Find out whether a or c has a greater depth in the DFS tree. Let’s say d = deepest(a, c). Color the edge d and parent[d] in red. This way we preserve the tree structure of the blue subgraph and decrement the number of connected components in the red subgraph.

If you have 3 back edges that do not form such a triangle/cycle. Just color them in red.

I highly suggest drawing a tree structure with back edges to understand it better.

C was so good :D

Very goood problems A, B, C!

How to solve C? Any hints

Look at number of index i, such that s[i]='(' and s[i+1]!=')'

or you can just count '))'s in $$$(s)$$$

I just use DSU.

How is its time complexity?

I don't calculate but get only 31 ms.

oh. that's fast. maybe the time complexity is $$$O(\text{fastenough})$$$.

DSU here how? If you could explain your thought process it would be helpful

I use DSU and at the last step count the different number or parent.If make count+1 at '(' and count-1 at ')' you will see that all i for 1 to 2*n where count =1 or count=0 are conected. I use this obserbation to solve. Sorry for my poor English.

Link every opening bracket to it's corresponding closing bracket. Whenever there is a ")(" in the string link those two brackets to handle the "()()" case where 1 needs to connect to 4.

the number of positions such that s[i] = s[i+1] = ')'

## F is shit.

Can't agree more.

Problem C was nice.

Wut it's easier than A, I'm mad because of B not saying it's hard but I'm so stupid and slow

me here, getting 150 on A :(((((((((((((((

Good sir, can I know how you made that observation for C?

I have never been so mad at a problem T-T

I noticed that () () () for example is all connected then I assumed that all of are disconnected and everytime i see () then this one is connected to another sequence

"I assumed that all of" you mean 'other' instead of 'of', right?

Also, Thanks but I don't know why I couldn't think of this. Would it be lack of practice of 1400-1500 rated problems?

It needs some luck i may haven't noticed it but if we are trained on graphs it won't be a problem

I see, thanks once again :D

I JUST use DSU and at the last step count the different number or parent.If make count+1 at '(' and count-1 at ')' you will see that all i for 1 to 2*n where count =1 or count=0 are conected. I use this obserbation to solve. Sorry for my poor English.

Thank you! I guess having a bit more knowledge of dsu might have helped ;(

It doesn't need any of that just a for loop would work

Yea true, what I actually meant was, maybe knowing dsu well might have given me a different perspective because I couldn't make that observation.

here's my crude drawing during contest for C, if it might help:

notice how the component count increases for every two decreasing (or increasing, as others pointed out) steps in a row.

Ohh nice, I will try this idea after few days and see if I can get the approach. Honestly, what I did was to notice the how deep the opening bracket's are going and counted them.

For eg. ( ( ( ) ) ), depth was 2.

And for this ( ( ) ) ( ( ) ), the depth for both first and second balanced sequence is 1. So, it becomes 2 and added 1 for the remaining component. And, it even passed for all given TCs, which made me believe theres something wrong in the code and not in observation.

I get riddled in my own ideas ;-;

What i thought is when the string is ((())) complete then the answer is n but the time you take out a () outside, a component decreases. So calculate no of instant parenthesis. We already have 1 instant parenthesis in the starting case. So answer will be n-(instant-1).

Ohh wow, this actually makes sense to me. Wish I would have thought a bit in this direction.

Alsoo, Congrats on specialist!!

i messed up preety badly at c

Problem F is boring and hard to implement :(

Any hints for D?

The connected components are trees.

Could you explain how to find spanning tree such that remaining 3 edges don't form a cycle( triangle )?

Random algorithm .

How to solve d with random algorithm

https://codeforces.com/blog/entry/106586?#comment-950389

Determenistic solution: Do a DFS, for every vertex record parent and depth in DFS tree, find the 3 additional edges. If the edges do form a cycle, take any of the 3 edges (let's say (u-v), where depth[u] > depth[v]) and replace it with edge (u-parent[u]).

Find any tree, then, if the remaining edges form a cycle, build a new tree with one edge of this cycle in it. It will guarantee that the new tree is ok.

Nope, for an example, in test case 2, you have one connected component and it is cycle.

I mean the connected components formed by same color.

Oh, okey when.

By the way , The $$$O(n\times \sqrt{nlog_2n})$$$ solution may be easier to implement.

FST now :(

hints for D?

I also failed it, my approach was this:

use union find to color all red without loops

at the end at most 3 blue edges will be left over

if those three blue edges happen to form a loop, swap the last blue edge with one of the red edges -> there will be no blue loop any more

my problem was I didn't know how to select the correct red edge to swap with, since for some red edges you will create a red-cycle

You have to check if the edge you are changing from blue -> red should not be between an ancestor and child in the red tree

Hint 1Note that for trees the answer is always n + 1 no matter what coloring you will use.

Hint 2Since graph is almost a tree (at most 3 more edges) think in terms of spanning trees.

Hint 3Consider two cases: if non-spanning tree don't form a triangle and when they do.

SolutionWhen $$$m \leq n + 1$$$ or $$$m = n + 2$$$ but the edges don't form a triangle, just color the whole spanning tree in blue and the non-spanning-tree edges in red. Otherwise, color two of the triangle edges blue and one red (call it $$$X$$$). Find the shortest paths IN THE SPANNING TREE from $$$X$$$ to the remaning vertex $$$v$$$ and colour neighboring edges that are on the path from $$$X$$$ endpoints to $$$v$$$ red. Everything else is blue (3 edges red, and remaning n — 1 blue). Draw some pictures to understand this idea better ;)

How is this optimal?

Each edge has to decrease number of either red components or blue components, and it can only decrese one of those by exactly 1. Thus, the optimal value is $$$2 * n - m$$$. In case of there not being a triangle we have exactly this value with the above method. In case of triangle also this value, but it's a little bit harder to see.

Defeated by the memory limit of problem F.

I wrote a segement tree without optimizing anything in the last minutes, using the memory of $$$O(n\log n)$$$. When I noticed the memory limits I didn't have time to make a change.

Same here. 256MB is unnecessarily too tight! Lose the chance to gain rating.

My ST table didn't get MLE.

However, I forgot to check whether there's a time which can reach n without passing any red lights and got WA on test 7. What a pity.

What is the expected time complexity for D?

Can we solve D by randomizing? My idea is you only need to randomize m — n + 1 edges, and check if any of the two sets create a cycle. m-n+1 is at most 3 so it should be fast enough?

The proof of correctness is that, if the graph is a tree, obviously it makes no difference how the edges are allocated. If we have extra edges, then every edges will reduce the connected component by 1, so we make a tree for the blue, and the rest for red, as long as they are both trees.

Make MST. Pick the $$$m-(n-1)$$$ edges not in MST. Only when $$$m = n+2$$$ can these edges create a cycle. If they do, then take one edge and transfer it to other color and the edge to its parent to current color, provided that the swapped edge is not in its subtree.

What is a 'connected component' in Problem D? In the first example, it says blue edges form 2 components, but I only see 1.

$$$1,2,3,5$$$ was one component, $$$4$$$ was the other

F is so annoying to implement (256MB memory limit makes it even worse!). But D and E are interesting.

## How to solve E? I thought its about calculating sums of $$$\frac{P^n_{2k}}{2^k k!}$$$ or smth for $$$1\leq k \leq \lfloor \frac{n}{2} \rfloor$$$

(sorry about the big text, things were becoming unreadable)

Yeah, the way you do that subproblem is by thinking about it conceptually instead of combinatorially. You're trying to find the ways to pair some elements from n things. This can be written as a

`dp[n] = dp[n-1] + (n-1)*dp[n-2]`

(you either pair the first element with something, or u don't), which can be precomputed before you do the loop over number of 4-cycles.Instead of permutations I thought about the problem with cycles (by building the graph of the permutation). To be almost perfect, cycles either have length 1 (pointing to itself) or length 2 (pointing to each other) or length 4, see picture. Rest was combinatorics. I will link my solution after the grading finished (was 7 minutes too late sadly...). I iterated over the amount of length-4 cycles.

OHHHHHHH. I was thinking about cycles too, in this process I came up with that formula. totally missed the length-4 cycles though. E was a good problem after all!

171160691.

First I calculated $$$P_i$$$ by this recursion: $$$P_0=1$$$, $$$P_1=1$$$, $$$P_i=P_{i-1}+(i-1)\cdot P_{i-2}$$$. This is the amount of partitions of $$$i$$$ elements into sets of size $$$1$$$ and $$$2$$$.

Then we calculate the following:

$$$\sum_{i=0}^{4i\le n} \binom{n-2i}{2i} (2i-1)!! \cdot 2^i P_{n-4i} $$$

The sum goes over the amount of length-$$$4$$$ cycles. The binomial calculates how many different distributions for $$$2i$$$ neighbouring 2-tile-pairs there are. The Double Factorial is the amount of combinations to form $$$4$$$-tile-cycles by choosing $$$2$$$ each from the previously distributed $$$2$$$-tile-pairs. Each $$$4$$$-tile cycle then can have 2 different orientations, so we need to multiply by $$$2^i$$$. Then $$$P_{n-4i}$$$ is the amount of combinations to form $$$1$$$ or $$$2$$$ tile pairs from the remaining nodes.

why getting TLE at problem B? someone help me

@u1602016 You are not decrementing variable t, so it continues to loop forever :)

oops sorry,last line just got canceled while copy pasting the code.still getting TLE

Can you share the your original submission link?

Yeah sure https://codeforces.com/contest/1726/submission/171148335

use sys module for input, here's my solution- https://codeforces.com/submissions/Varad2002

thanks a lot @varad2002

You’re welcome :)

Same problem. TLE with Python/PyPy3 (pretest 4), OK with C++ (system testing). Slow input/output in Python, maybe.

Is it possible to get pretest 4 for experiments?

Submission https://codeforces.com/contest/1726/submission/171110352 Room (with the other submissions with slightly smodified input/output) https://codeforces.com/contest/1726/room/464

The way it works is that the

`input`

function in Python calls`sys.stdout.flush`

. So everytime you call`input`

you flush stdout, which can be really slow. The way to get around this is to use`sys.stdin.readline`

instead of`input`

. Worth noting is that`input`

internally calls`sys.stdin.readline`

.If you take a look at C++ submissions will probably see

`cin.tie(0)`

, which is how you turn off this very same feature in C++.Knowing how to avoid flushing can be very useful in general. For example your (savsmail) C++ submission speeds up by a factor of 16 by avoiding to flush 171253842 .

Wow, that explains a lot. Thank you!

@u1602016 yes, the problem is with the slow I/O in python as it takes 5-10 times more to run compare to c/c++.

For the cross-check part I have also converted your solution in c++, and tried to submit, and it works perfectly fine within time(561ms, note: your solution is correct) here

Solution???you could either use the fast I/O for python while doing CP (as @Varad2002 mentioned) or you could use c++ for CP (I would personally recommend this, as I have experienced the situation where my JAVA code throws TLE (even after using fast I/O) whereas same algorithm works within time limit in C++)thanks a lot for your suggestions.@Nikhil19259

I couldn't enjoy this contest :(, might be very clever problems.

hint for c?

)(

Spoiler 1Each component is enclosed in a corresponding Open-Close-bracket

Spoiler 2You just need to count how many bracketpairs have non-empty content inside them.

SolutionCount the occurences of

`((`

and add one. 171090971Was this easy to catch this observation as i tried to solve trivally by making DSU but not able to do it

Phew, well, depends on your understanding of easy. As I was solving the problem, the Spoilers/Solution which I posted reconstruct my thought process. And I'd say I have noticed them pretty quickly. I didn't even consider building the graph explicitly (and using DSU or similar) yet.

I guess the first thing you want to do is find observations about the structure of the problem. Only if you can't find any structure, you take out the harder techniques. And then try to find structure on the harder techniques. And

alwaysuser pen and paper!Any hints for E?

Look at cycle decomposition of $$$p$$$.

Really nice problems ! Keep it up authors ^_^

D: randomly color all edges. Break if both colors do not form cycles.

We tried our best to break such solutions, but it seems it was not enough :(

I would like to know how to break (or attempt to) these kind of solutions.

My attempts were primarily focused on maximising the number of tests, with $$$n+2$$$ edges in every case, with graphs of ~1e3 (I think?) size. So mostly randomisation. I didn't have much hope of blocking everything because of the small number of extra edges compared to the number of possible spanning trees. If anyone has better ideas, I would be very interested to know...

Break? Isn't that correct?

Some of our testers' solutions using the same idea didn't pass, so I assumed it is incorrect.

Of course, if the number of iterations are sufficient it should pass with high probability. Admittedly, I did not do a thorough analysis of a randomised solution because the model solution does not use randomisation.

Well, this one looks way simpler than anything with some logic inside. I'm talking about the version with a guess that we can achieve maximum possible score ($$$2n - m$$$) and we repeat until we succeed.

The intuition for me was that if it's anyhow hard/not straightforward to color the edges so that there are no cycles, these additional edges (yes, it's relative which ones are additional, but I'm talking about kind of a part of the graph in which something interesting happens) have to be close to each other, so the "hard" part of the graph is small, so there are not so many possibilities and we will find good one quickly.

I did this: Colored red if nodes are in different disjoint sets, blue otherwise. However, for m=n+2, there is a possibility of a cycle of 3 formed.

In that case, I made this observation: Suppose the 3-cycle is ABC then there is a node in A,B,C say A, such that AB and AC are connected using red. Then just take the last edge in the AB red path and make it blue, and make AB(blue) as red. It can be seen that this removes cycles in both red and blue.

Why the downvotes?

Really enjoyed the contest, especially because all the questions were not only mathematics based. Also, waiting to submit D, solved just after the time ended. Same thing happening for the second time :')

D accepted in 1 go after the contest :')

Code of Problem C <<< Solving Problem C

wait what the heck

SpoilerI had to gain only 2 points to reach CM, and solved D on paper, started coding it more than 1 hr before the end, but in the end got TL 9 because i use maps which work slow. But my idea was fully correct. Now end up losing 7 points, and only 9 will be left before CM. this is fucking disqusting, it will be already 5th time when i have <10 points bafore CM!

Oh :/

Isn't $$$1.205710448301$$$ the answer for the first sample in H? I've checked that in geogebra and it seems to be so.

Are you sure it's the first sample? I mean to say, you don't mean the second sample by any chance, right?

I think so, here's the geogebra link: we are looking for the area bounded by $$$AB$$$, $$$AC$$$, and the circle passing through H and center O.

The area is bounded by a curve, but not an arc

If you are using a circle as an envelope for the red part, that's incorrect. The curve is not a circle.

Check this: Click here

Oh I see, thanks

My solution to problem C passsed system tests, Even though I am pretty sure its so different from the intended solution

I solved with a sparse table and DSU, I am interested to know if the solution is correct and actually fast enough (I am 99% sure its fast enough but I don't know about its validity)

Can Someone Verify?

Submission : https://codeforces.com/contest/1726/submission/171128759

do note that a lot of the faster ones (usually in Div1) solved it with DSU

I too solved it using seg tree and DSU.

Back to back bad performance from me, hope I can bounce back soon. People who are in a phase like me, "This shall pass too" :)

Check out this ridiculously easy solution for C: 171112911 (don't pay attention to that

`bin_power_of_two`

function — I forgot to delete it)The interesting thing is that I haven't found any of the red guys yet who would have approached the problem the same way.

yes, because for them, the simpler (in the sense of observation) DSU implementation came first in mind.

Could you explain the DSU solution?

they just constructed the DSU like how they would do a normal graph and counted the components like how they would do a normal graph. only tricky part is what to merge and that's it

171072071

ty

literally

`('('+input()+')').count('))')`

in python.Absolutely

Even this works

These unfamiliar names kinda throw me off the track. Wish we could use simpler names

If the current problem C(bracket sequence) were problem A, would it have been as difficult as it is now?

My exactly same code got ac after system test

AC code : https://codeforces.com/contest/1726/submission/171155848 TLE code : https://codeforces.com/contest/1726/submission/171146706

code was same compiler was same.... Just for system test load my code got tle...Please @authors do something about this...MikeMirzayanov please do something

code was same, compiler was same, the difference is the seed

what is seed?

the seed is the value an RNG uses to start with. in your code this is

`chrono::steady_clock::now().time_since_epoch().count()`

, which varies by when the submission is run.I haven't used rng anywhere in my code...so it was not my fault at all.

oh, didn't notice it. turns out that your code barely just fit into the TL. I think you would've gotten hacked if this happened during competition, though.

can you do something ? Newtech66

Do you really think 1996ms solution for 2s problem is correct?

How should I calculate this in contest time? Pretests got passed and even system tests too. But the judge load caused me TLE. Whats my fault here?

On A, I saw a solution which initializes the answer with $$$|a_n-a_1|$$$ and, while I could've simply made a test case which breaks that solution very obviously, I instead decided to make fun of the pretests by hacking it with a test case of 50 tests with $$$n$$$ being a random number between $$$1$$$ and $$$6$$$ and $$$a_i$$$ being random numbers between $$$1$$$ and $$$10$$$. While most obviously wrong solutions like the one I hacked FSTd due to hacks specifically made against such solutions, this test case of 50 random tests also unintentionally caused FST of almost 100 solutions that don't work if a test with $$$n=1$$$ has a test right before it such that $$$a_2$$$ of the test right before the test with $$$n=1$$$ is greater than $$$a_1$$$ of the $$$n=1$$$ test.

LOL that is awesome. this is a legend, as you are too

Why is the D's constraints so big? I can't understand why sum of N is 1e6. It is hard for python... (I passed pretest but got TLE on system test)

It was because we were attempting to block randomised solutions. I'm sorry about the situation with Python being too slow >_<

I caught TL52 on the task D (https://codeforces.com/contest/1726/submission/171125630). After contest I sent the same solution again and it got Accepted (https://codeforces.com/contest/1726/submission/171157258). How did it happen? (>2 seconds vs 1.7 seconds)

I didn't know randomized wasn't intended. Is this hackable? https://codeforces.com/contest/1726/submission/171121286

Done C with DSU... how u guys come up with so easy solution...

observation, but observation isn't all so easy. observation comes along with experience as well

I am not very comfortable with these graph problem as you can see from my rating that i havent encountered many graph problems till now . Do these graph problem usually have these kind observation asking wrt C

And please guide my what i should do to reach specialist as while practicing i am comfortable upto *1500 problems but during contest i keep on getting stucked on problem C which are usually in the range of *1400 to *1600

So am I! This was a different case though. I just drew some parentheses strings with pen and paper, like '(' = rising slope, ')' = falling slope. Then I connected what I needed to connect based on the statements. After that, it was a long phase of thinking, thinking hard. Usually time for thinking > time for coding, this is very normal. Accepting this is one crucial step to Specialist, imho.

Ratings updated preliminarily. We will remove cheaters and update the ratings again soon!

Got Accept with FST code for D.

Why is this so slow? It's just O(NlogN) and N <= 1e6

Can someone explain?

https://codeforces.com/contest/1726/submission/171146699

https://codeforces.com/contest/1726/submission/171158814

Same happened to me and costed me losing CM. @admins should look at this.

maybe try having a function fix(pair p) that returns a sorted pair(smaller element before larger element). idk if it fixes your solution but it did work for my friend

Btw, I FST for D, but when I try to resubmit the exact same code, it passes (barely). Is it possible to reverse the FST?

Faced the same issue. Not sure if they will take it into consideration, but I hope they do.

Hello everyone, I'm a newbie participant. On problem A, although it runs perfectly on my computer, I got runtime error. How Can I fix this problem? I copy my code below. Thank you. (I use Microsoft Visual Studio BTW)

you need to increase the size of array. notice n<=2000 but you have declared an array of size 1000 only

After decades of the contest, Benq is on the top! Happy to see this (❁´◡`❁)

He didn't get the rating, though.

Just a shit

Before I sleep,I saw my name turn green..What F???

Because F!!!

lol everyone is downvoting this post and the editorial.

And even the comments, XD.

SpoilerThe coding club of IIT KGP is gonna face a lot of disrepute now becoz of you guys

Exactly , they may set 7 problems instead of 8 and time for 2 Hours would be better.

A sincere request to everyone!Newtech66 and anubhavdhar have worked really hard for this round and they weren't aware of what happened in problem F. If you don't like the problems(except F), feel free to downvote the blog, but please refrain from doing so if this is the only reason! It would be very unfair and unjustice with them :(

This is bullshit. I support downvoting the blog post because of F, it's high time all the authors start asking their co-authors how they developed the problems, so situations like this don't repeat. Every author of the round should be held equally accountable for each problem, and the round in general.

where is my ratings for this contest.

unrated because of a problem dupe

ppl downvotes for literally no reason lol

Why this contest is Unrated?

Problem F was a stolen problem.

The discovery

The announcement

my rating where is gone?!! i just reached pupil pls any information?

https://codeforces.com/blog/entry/106700

This is has happened 2nd or 3rd time, that contest has been declared unrated due to picking problems from other platforms. Please make sure that problems are original beforehand. There goes a lot of effort in solving them and then finding that contest has been declared unrated just breaks the heart :( .

cope kro