F is shit.

2 years ago, # ^ |

+65

when asking for downvotes and asking for upvotes are unsurprisingly in the same SCC

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bycicle

2 years ago, # ^ |

-48

As a participant, I will have participated.

Spoiler for upvotes:

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2 years ago, # ^ |

+12

you've reposted in the wrong connected component

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User_Carrot

2 years ago, # ^ |

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

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Greedious

2 years ago, # ^ |

← Rev. 2 →

+60

posting mind blowing meme to get upvotes :

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shivansh1102

2 years ago, # |

+60

As the leamder

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absolute-mess

2 years ago, # ^ |

+45

Domne

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anubhavdhar

2 years ago, # |

+52

As an author

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Tom66

2 years ago, # |

+30

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

When will scoring distribution come out?

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physics0523

2 years ago, # |

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

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glebustim

2 years ago, # |

+30

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

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BARBARIANNNNN

2 years ago, # |

-10

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

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DitaMirika

2 years ago, # |

← Rev. 2 →

+15

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

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2 years ago, # ^ |

+14

As an author, me too.

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DitaMirika

2 years ago, # ^ |

Looking forward to it :p

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2 years ago, # |

-262

As an author

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2 years ago, # ^ |

-8

omg that thumbs up emoji looks fabulous

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2 years ago, # ^ |

-21

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)

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2 years ago, # ^ |

-22

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

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2 years ago, # ^ |

-251

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

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2 years ago, # ^ |

+76

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...

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flakes24

2 years ago, # ^ |

got downvoted for being right

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Jynxzi

2 years ago, # ^ |

now this is a comeback

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2 years ago, # ^ |

help me, this comeback has reversed somehow

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2 years ago, # ^ |

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!

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PolarFlea

2 years ago, # ^ |

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

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AntiXoR

2 years ago, # |

← Rev. 2 →

IIT KGP !orz

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cf-coder

2 years ago, # ^ |

What is IIT KGP?

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drakeramoray27

2 years ago, # ^ |

+10

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

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ZyadFva

2 years ago, # |

As a pupil contestant

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Hamroz_JONN

2 years ago, # ^ |

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

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Echp0chmak

2 years ago, # ^ |

and what's so wrong with it?

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ZyadFva

2 years ago, # ^ |

-8

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

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whynesspower

2 years ago, # ^ |

why is it unrated?

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ZyadFva

2 years ago, # ^ |

one of problems was copied from another judge.

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allal

2 years ago, # |

I hope I can do at least the first one

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Omar_Hafez

2 years ago, # |

+91

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2 years ago, # ^ |

+38

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

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

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2 years ago, # ^ |

first time ?

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tiasmondal

2 years ago, # |

+50

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

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SalmaHatem_

2 years ago, # |

← Rev. 2 →

+10

I love div.1 + div.2 contests
Good luck for everyone!!
Hoping for a nice problems❤️

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nowadays-

2 years ago, # ^ |

Good luck

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islam231

2 years ago, # |

← Rev. 2 →

Is it rated for DIV 2 ?

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2 years ago, # ^ |

+14

Yes, it is rated for all divisions.

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pirate_hunter_zoro

2 years ago, # ^ |

not any more xD

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Noobish_Monk

2 years ago, # |

Hoping to solve 4 problems, would be great

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2 years ago, # ^ |

-259

Wishing you luck, to solve even more

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nowadays-

2 years ago, # ^ |

Hope you can solve more problems too

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Retired_Snow

2 years ago, # |

Hope to become as soon as possible Candidate Master.

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2 years ago, # ^ |

-266

All the best

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cuzperf

2 years ago, # |

As a xxx

Here is: hided content

The above comment is edited as follows:

<div class="spoiler">
  <b class="spoiler-title">As a xxx</b>
  <div class="spoiler-content" style="display: none;">
    <p>Give me xxx</p>
  </div>
</div>

Here is: <span style="color: white;">hided content</span>

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thelastdinner

2 years ago, # ^ |

+32

However, A new comment will be shown in blue background

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Abito

2 years ago, # ^ |

hidden content*

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Allen123

2 years ago, # |

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

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2 years ago, # ^ |

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

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pigmike

2 years ago, # ^ |

+23

god, you're everywhere

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Silly__coder__

2 years ago, # |

-12

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

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hackerbhaiya

2 years ago, # ^ |

+22

It's IND vs SL!

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mr-strange

2 years ago, # |

All the Best EveryOne!

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nowadays-

2 years ago, # |

Good luck to all

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Ahmed_AboRadwan

2 years ago, # |

Sometimes you win, sometimes you learn

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2 years ago, # |

We just got trolled

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2 years ago, # |

How to solve D?

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dulatcodes

2 years ago, # ^ |

← Rev. 2 →

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.

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Saucemaster102

2 years ago, # |

+17

C was so good :D

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AlexMari

2 years ago, # |

Very goood problems A, B, C!

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Ayanokouji

2 years ago, # |

How to solve C? Any hints

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BULLMCHOW

2 years ago, # ^ |

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

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2 years ago, # ^ |

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

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2 years ago, # ^ |

I just use DSU.

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2 years ago, # ^ |

← Rev. 2 →

How is its time complexity?

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2 years ago, # ^ |

I don't calculate but get only 31 ms.

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2 years ago, # ^ |

← Rev. 2 →

+22

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

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Ayanokouji

2 years ago, # ^ |

← Rev. 3 →

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

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2 years ago, # ^ |

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.

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robostac

2 years ago, # ^ |

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.

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doraty228

2 years ago, # ^ |

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

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hzr

2 years ago, # |

+233

F is shit.

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Lightnessqwq

2 years ago, # ^ |

+73

Can't agree more.

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2 years ago, # |

Problem C was nice.

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2 years ago, # ^ |

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

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2 years ago, # ^ |

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

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2 years ago, # ^ |

Good sir, can I know how you made that observation for C?
I have never been so mad at a problem T-T

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2 years ago, # ^ |

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

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2 years ago, # ^ |

"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?

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2 years ago, # ^ |

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

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2 years ago, # ^ |

I see, thanks once again :D

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2 years ago, # ^ |

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.

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2 years ago, # ^ |

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

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2 years ago, # ^ |

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

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2 years ago, # ^ |

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.

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2 years ago, # ^ |

← Rev. 2 →

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.

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2 years ago, # ^ |

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 ;-;

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strgglr

2 years ago, # ^ |

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).

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2 years ago, # ^ |

Ohh wow, this actually makes sense to me. Wish I would have thought a bit in this direction.
Alsoo, Congrats on specialist!!

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mukulrawat_03

2 years ago, # ^ |

i messed up preety badly at c

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2 years ago, # |

+40

Problem F is boring and hard to implement :(

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shauryasinha020

2 years ago, # ^ |

Any hints for D?

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2 years ago, # ^ |

The connected components are trees.

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drdilyor

2 years ago, # ^ |

← Rev. 2 →

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

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2 years ago, # ^ |

+30

Random algorithm .

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nidaboge

2 years ago, # ^ |

How to solve d with random algorithm

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

2 years ago, # ^ |

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ikaurov

2 years ago, # ^ |

+34

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]).

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thatsramen

2 years ago, # ^ |

+32

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.

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BULLMCHOW

2 years ago, # ^ |

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

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2 years ago, # ^ |

I mean the connected components formed by same color.

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BULLMCHOW

2 years ago, # ^ |

← Rev. 2 →

Oh, okey when.

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2 years ago, # ^ |

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

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2 years ago, # ^ |

FST now :(

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shauryasinha020

2 years ago, # |

hints for D?

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z1bbo

2 years ago, # ^ |

← Rev. 3 →

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

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shadow9236

2 years ago, # ^ |

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

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Wielomian

2 years ago, # ^ |

Hint 1

Hint 2

Hint 3

Solution

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2 years ago, # ^ |

How is this optimal?

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Wielomian

2 years ago, # ^ |

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.

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gyh20

2 years ago, # |

← Rev. 2 →

+34

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.

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dl720125

2 years ago, # ^ |

+10

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

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tzc_wk

2 years ago, # ^ |

+20

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.

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__hermit__

2 years ago, # |

What is the expected time complexity for D?

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algoboi

2 years ago, # |

← Rev. 2 →

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.

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shadow9236

2 years ago, # ^ |

+12

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.

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koqo

2 years ago, # |

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

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2 years ago, # ^ |

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

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feecIe6418

2 years ago, # |

+41

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

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2 years ago, # ^ |

← Rev. 2 →

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)

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bajablast

2 years ago, # ^ |

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.

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2 years ago, # ^ |

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.

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2 years ago, # ^ |

← Rev. 2 →

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!

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2 years ago, # ^ |

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.

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2 years ago, # |

← Rev. 7 →

why getting TLE at problem B? someone help me

t = int(input())
while t>=1:
    n,m = map(int,input().split())
    if m<n:
        print("No")
 
    elif n%2 == 0 and m%2 !=0 :
        print("No")
 
    else:
        ans = []
        if n%2 !=0:
            ans = [1] * (n-1)
            a = m - (n-1)
            ans.append(a)
        else:
            ans = [1] * (n-2)
            a = (m - (n-2))//2
            ans.append(a)
            ans.append(a)
 
        print("Yes")
        for i in range(len(ans)):
            if i == len(ans)-1:
                print(ans[i])
            else: print(ans[i],end= " ")
    t-=1

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Nikhil19259

2 years ago, # ^ |

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

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2 years ago, # ^ |

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

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Nikhil19259

2 years ago, # ^ |

Can you share the your original submission link?

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2 years ago, # ^ |

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

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Varad

2 years ago, # ^ |

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

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2 years ago, # ^ |

thanks a lot @varad2002

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Varad

2 years ago, # ^ |

You’re welcome :)

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savsmail

2 years ago, # ^ |

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

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pajenegod

2 years ago, # ^ |

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 .

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savsmail

2 years ago, # ^ |

Wow, that explains a lot. Thank you!

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Nikhil19259

2 years ago, # ^ |

@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++)

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2 years ago, # ^ |

thanks a lot for your suggestions.@Nikhil19259

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d_sqaure

2 years ago, # |

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

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Rikitexh

2 years ago, # |

hint for c?

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Citypop

2 years ago, # ^ |

)(

→ Reply

2 years ago, # ^ |

+12

Spoiler 1

Spoiler 2

Solution

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faizanhussain2310

2 years ago, # ^ |

Was this easy to catch this observation as i tried to solve trivally by making DSU but not able to do it

→ Reply

2 years ago, # ^ |

← Rev. 2 →

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 always user pen and paper!

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Theo830

2 years ago, # |

Any hints for E?

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bajablast

2 years ago, # ^ |

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

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TanSinTheta

2 years ago, # |

Really nice problems ! Keep it up authors ^_^

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rgnerdplayer

2 years ago, # |

+16

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

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2 years ago, # ^ |

+20

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

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rgnerdplayer

2 years ago, # ^ |

+13

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

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2 years ago, # ^ |

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...

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Radewoosh

2 years ago, # ^ |

+21

Break? Isn't that correct?

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2 years ago, # ^ |

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.

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Radewoosh

2 years ago, # ^ |

+17

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.

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2 years ago, # ^ |

← Rev. 2 →

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.

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2 years ago, # ^ |

Why the downvotes?

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2 years ago, # |

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 :')

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2 years ago, # ^ |

+16

D accepted in 1 go after the contest :')

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2 years ago, # |

+16

Code of Problem C <<< Solving Problem C

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2 years ago, # |

-13

wait what the heck

Spoiler

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M_bolshakov

2 years ago, # |

-14

I 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!

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palindroam

2 years ago, # ^ |

Oh :/

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sstrong

2 years ago, # |

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

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2 years ago, # ^ |

← Rev. 2 →

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

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sstrong

2 years ago, # ^ |

← Rev. 2 →

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.

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Artyom123

2 years ago, # ^ |

+26

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

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2 years ago, # ^ |

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

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sstrong

2 years ago, # ^ |

Oh I see, thanks

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Yahia_Emara

2 years ago, # |

← Rev. 2 →

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.net/contest/1726/submission/171128759

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2 years ago, # ^ |

-8

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

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dream7

2 years ago, # ^ |

I too solved it using seg tree and DSU.

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Cooper1214

2 years ago, # |

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" :)

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2 years ago, # |

← Rev. 2 →

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)

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2 years ago, # ^ |

The interesting thing is that I haven't found any of the red guys yet who would have approached the problem the same way.

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2 years ago, # ^ |

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

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2 years ago, # ^ |

Could you explain the DSU solution?

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2 years ago, # ^ |

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

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rgnerdplayer

2 years ago, # ^ |

-8

171072071

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2 years ago, # ^ |

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2 years ago, # ^ |

literally ('('+input()+')').count('))') in python.

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2 years ago, # ^ |

Absolutely

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2 years ago, # ^ |

Even this works

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2 years ago, # ^ |

    ll ans=1;
    forsn(i,1,2*n){
        if(s[i]=='(' && s[i-1]=='(') ans++;
    }

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7_7

2 years ago, # |

-10

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

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2 years ago, # |

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

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2 years ago, # |

My exactly same code got ac after system test

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

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2 years ago, # ^ |

-10

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

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2 years ago, # ^ |

what is seed?

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2 years ago, # ^ |

← Rev. 2 →

-10

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.

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2 years ago, # ^ |

I haven't used rng anywhere in my code...so it was not my fault at all.

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2 years ago, # ^ |

-10

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.

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2 years ago, # ^ |

can you do something ? Newtech66

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MrOtter

2 years ago, # ^ |

+15

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

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2 years ago, # ^ |

-7

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?

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dorijanlendvaj

2 years ago, # |

← Rev. 2 →

+116

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.

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2 years ago, # ^ |

-13

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

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toam

2 years ago, # |

+23

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)

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2 years ago, # ^ |

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

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FynjyBath

2 years ago, # ^ |

← Rev. 2 →

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

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Kyou_mo_kawaii

2 years ago, # ^ |

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

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shivam_360

2 years ago, # |

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

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2 years ago, # ^ |

-13

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

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faizanhussain2310

2 years ago, # ^ |

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

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https://codeforces.net/contest/1726/submission/171146699
https://codeforces.net/contest/1726/submission/171158814

2 years ago, # ^ |

I am not very comfortable with these graph problem

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.

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geranazavr555

2 years ago, # |

+10

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

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ganon

2 years ago, # |

← Rev. 2 →

Got Accept with FST code for D.
Why is this so slow? It's just O(NlogN) and N <= 1e6
Can someone explain?

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