F is shit.

2 года назад, # ^ |

+65

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

→ Ответить

bycicle

2 года назад, # ^ |

-48

As a participant, I will have participated.

Spoiler for upvotes:

→ Ответить

2 года назад, # ^ |

+12

you've reposted in the wrong connected component

→ Ответить

User_Carrot

2 года назад, # ^ |

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

→ Ответить

Greedious

2 года назад, # ^ |

← Rev. 2 →

+60

posting mind blowing meme to get upvotes :

→ Ответить

shivansh1102

2 года назад, # |

+60

As the leamder

→ Ответить

absolute-mess

2 года назад, # ^ |

+45

Domne

→ Ответить

anubhavdhar

2 года назад, # |

+52

As an author

→ Ответить

Tom66

2 года назад, # |

+30

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

When will scoring distribution come out?

→ Ответить

physics0523

2 года назад, # |

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

→ Ответить

glebustim

2 года назад, # |

+30

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

→ Ответить

BARBARIANNNNN

2 года назад, # |

-10

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

→ Ответить

DitaMirika

2 года назад, # |

← Rev. 2 →

+15

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

→ Ответить

2 года назад, # ^ |

+14

As an author, me too.

→ Ответить

DitaMirika

2 года назад, # ^ |

Looking forward to it :p

→ Ответить

2 года назад, # |

-262

As an author

→ Ответить

2 года назад, # ^ |

-8

omg that thumbs up emoji looks fabulous

→ Ответить

2 года назад, # ^ |

-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)

→ Ответить

2 года назад, # ^ |

-22

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

→ Ответить

2 года назад, # ^ |

-251

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

→ Ответить

2 года назад, # ^ |

+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...

→ Ответить

flakes24

2 года назад, # ^ |

got downvoted for being right

→ Ответить

Jynxzi

2 года назад, # ^ |

now this is a comeback

→ Ответить

2 года назад, # ^ |

help me, this comeback has reversed somehow

→ Ответить

2 года назад, # ^ |

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!

→ Ответить

PolarFlea

2 года назад, # ^ |

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

→ Ответить

AntiXoR

2 года назад, # |

← Rev. 2 →

IIT KGP !orz

→ Ответить

cf-coder

2 года назад, # ^ |

What is IIT KGP?

→ Ответить

drakeramoray27

2 года назад, # ^ |

+10

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

→ Ответить

ZyadFva

2 года назад, # |

As a pupil contestant

→ Ответить

Hamroz_JONN

2 года назад, # ^ |

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

→ Ответить

Echp0chmak

2 года назад, # ^ |

and what's so wrong with it?

→ Ответить

ZyadFva

2 года назад, # ^ |

-8

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

→ Ответить

whynesspower

2 года назад, # ^ |

why is it unrated?

→ Ответить

ZyadFva

2 года назад, # ^ |

one of problems was copied from another judge.

→ Ответить

allal

2 года назад, # |

I hope I can do at least the first one

→ Ответить

Omar_Hafez

2 года назад, # |

+91

→ Ответить

2 года назад, # ^ |

+38

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

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

→ Ответить

2 года назад, # ^ |

first time ?

→ Ответить

tiasmondal

2 года назад, # |

+50

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

→ Ответить

SalmaHatem_

2 года назад, # |

← Rev. 2 →

+10

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

→ Ответить

nowadays-

2 года назад, # ^ |

Good luck

→ Ответить

islam231

2 года назад, # |

← Rev. 2 →

Is it rated for DIV 2 ?

→ Ответить

2 года назад, # ^ |

+14

Yes, it is rated for all divisions.

→ Ответить

pirate_hunter_zoro

2 года назад, # ^ |

not any more xD

→ Ответить

Noobish_Monk

2 года назад, # |

Hoping to solve 4 problems, would be great

→ Ответить

2 года назад, # ^ |

-259

Wishing you luck, to solve even more

→ Ответить

nowadays-

2 года назад, # ^ |

Hope you can solve more problems too

→ Ответить

Retired_Snow

2 года назад, # |

Hope to become as soon as possible Candidate Master.

→ Ответить

2 года назад, # ^ |

-266

All the best

→ Ответить

cuzperf

2 года назад, # |

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>

→ Ответить

thelastdinner

2 года назад, # ^ |

+32

However, A new comment will be shown in blue background

→ Ответить

Abito

2 года назад, # ^ |

hidden content*

→ Ответить

Allen123

2 года назад, # |

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

→ Ответить

2 года назад, # ^ |

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

→ Ответить

pigmike

2 года назад, # ^ |

+23

god, you're everywhere

→ Ответить

Silly__coder__

2 года назад, # |

-12

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

→ Ответить

hackerbhaiya

2 года назад, # ^ |

+22

It's IND vs SL!

→ Ответить

mr-strange

2 года назад, # |

All the Best EveryOne!

→ Ответить

nowadays-

2 года назад, # |

Good luck to all

→ Ответить

Ahmed_AboRadwan

2 года назад, # |

Sometimes you win, sometimes you learn

→ Ответить

2 года назад, # |

We just got trolled

→ Ответить

2 года назад, # |

How to solve D?

→ Ответить

dulatcodes

2 года назад, # ^ |

← 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.

→ Ответить

Saucemaster102

2 года назад, # |

+17

C was so good :D

→ Ответить

AlexMari

2 года назад, # |

Very goood problems A, B, C!

→ Ответить

Ayanokouji

2 года назад, # |

How to solve C? Any hints

→ Ответить

BULLMCHOW

2 года назад, # ^ |

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

→ Ответить

2 года назад, # ^ |

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

→ Ответить

2 года назад, # ^ |

I just use DSU.

→ Ответить

2 года назад, # ^ |

← Rev. 2 →

How is its time complexity?

→ Ответить

2 года назад, # ^ |

I don't calculate but get only 31 ms.

→ Ответить

2 года назад, # ^ |

← Rev. 2 →

+22

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

→ Ответить

Ayanokouji

2 года назад, # ^ |

← Rev. 3 →

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

→ Ответить

2 года назад, # ^ |

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.

→ Ответить

robostac

2 года назад, # ^ |

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.

→ Ответить

doraty228

2 года назад, # ^ |

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

→ Ответить

hzr

2 года назад, # |

+233

F is shit.

→ Ответить

Lightnessqwq

2 года назад, # ^ |

+73

Can't agree more.

→ Ответить

2 года назад, # |

Problem C was nice.

→ Ответить

2 года назад, # ^ |

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

→ Ответить

2 года назад, # ^ |

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

→ Ответить

2 года назад, # ^ |

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

→ Ответить

2 года назад, # ^ |

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

→ Ответить

2 года назад, # ^ |

"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?

→ Ответить

2 года назад, # ^ |

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

→ Ответить

2 года назад, # ^ |

I see, thanks once again :D

→ Ответить

2 года назад, # ^ |

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.

→ Ответить

2 года назад, # ^ |

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

→ Ответить

2 года назад, # ^ |

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

→ Ответить

2 года назад, # ^ |

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.

→ Ответить

2 года назад, # ^ |

← 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.

→ Ответить

2 года назад, # ^ |

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 ;-;

→ Ответить

strgglr

2 года назад, # ^ |

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).

→ Ответить

2 года назад, # ^ |

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

→ Ответить

mukulrawat_03

2 года назад, # ^ |

i messed up preety badly at c

→ Ответить

2 года назад, # |

+40

Problem F is boring and hard to implement :(

→ Ответить

shauryasinha020

2 года назад, # ^ |

Any hints for D?

→ Ответить

2 года назад, # ^ |

The connected components are trees.

→ Ответить

drdilyor

2 года назад, # ^ |

← Rev. 2 →

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

→ Ответить

2 года назад, # ^ |

+30

Random algorithm .

→ Ответить

nidaboge

2 года назад, # ^ |

How to solve d with random algorithm

→ Ответить

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

2 года назад, # ^ |

→ Ответить

ikaurov

2 года назад, # ^ |

+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]).

→ Ответить

thatsramen

2 года назад, # ^ |

+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.

→ Ответить

BULLMCHOW

2 года назад, # ^ |

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

→ Ответить

2 года назад, # ^ |

I mean the connected components formed by same color.

→ Ответить

BULLMCHOW

2 года назад, # ^ |

← Rev. 2 →

Oh, okey when.

→ Ответить

2 года назад, # ^ |

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

→ Ответить

2 года назад, # ^ |

FST now :(

→ Ответить

shauryasinha020

2 года назад, # |

hints for D?

→ Ответить

z1bbo

2 года назад, # ^ |

← 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

→ Ответить

shadow9236

2 года назад, # ^ |

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

→ Ответить

Wielomian

2 года назад, # ^ |

Hint 1

Hint 2

Hint 3

Solution

→ Ответить

2 года назад, # ^ |

How is this optimal?

→ Ответить

Wielomian

2 года назад, # ^ |

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.

→ Ответить

gyh20

2 года назад, # |

← 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.

→ Ответить

dl720125

2 года назад, # ^ |

+10

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

→ Ответить

tzc_wk

2 года назад, # ^ |

+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.

→ Ответить

__hermit__

2 года назад, # |

What is the expected time complexity for D?

→ Ответить

algoboi

2 года назад, # |

← 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.

→ Ответить

shadow9236

2 года назад, # ^ |

+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.

→ Ответить

koqo

2 года назад, # |

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

→ Ответить

2 года назад, # ^ |

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

→ Ответить

feecIe6418

2 года назад, # |

+41

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

→ Ответить

2 года назад, # ^ |

← 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)

→ Ответить

bajablast

2 года назад, # ^ |

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.

→ Ответить

2 года назад, # ^ |

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.

→ Ответить

2 года назад, # ^ |

← 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!

→ Ответить

2 года назад, # ^ |

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.

→ Ответить

2 года назад, # |

← 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

→ Ответить

Nikhil19259

2 года назад, # ^ |

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

→ Ответить

2 года назад, # ^ |

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

→ Ответить

Nikhil19259

2 года назад, # ^ |

Can you share the your original submission link?

→ Ответить

2 года назад, # ^ |

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

→ Ответить

Varad

2 года назад, # ^ |

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

→ Ответить

2 года назад, # ^ |

thanks a lot @varad2002

→ Ответить

Varad

2 года назад, # ^ |

You’re welcome :)

→ Ответить

savsmail

2 года назад, # ^ |

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

→ Ответить

pajenegod

2 года назад, # ^ |

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 .

→ Ответить

savsmail

2 года назад, # ^ |

Wow, that explains a lot. Thank you!

→ Ответить

Nikhil19259

2 года назад, # ^ |

@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++)

→ Ответить

2 года назад, # ^ |

thanks a lot for your suggestions.@Nikhil19259

→ Ответить

d_sqaure

2 года назад, # |

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

→ Ответить

Rikitexh

2 года назад, # |

hint for c?

→ Ответить

Citypop

2 года назад, # ^ |

)(

→ Ответить

2 года назад, # ^ |

+12

Spoiler 1

Spoiler 2

Solution

→ Ответить

faizanhussain2310

2 года назад, # ^ |

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

→ Ответить

2 года назад, # ^ |

← 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!

→ Ответить

Theo830

2 года назад, # |

Any hints for E?

→ Ответить

bajablast

2 года назад, # ^ |

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

→ Ответить

TanSinTheta

2 года назад, # |

Really nice problems ! Keep it up authors ^_^

→ Ответить

rgnerdplayer

2 года назад, # |

+16

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

→ Ответить

2 года назад, # ^ |

+20

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

→ Ответить

rgnerdplayer

2 года назад, # ^ |

+13

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

→ Ответить

2 года назад, # ^ |

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...

→ Ответить

Radewoosh

2 года назад, # ^ |

+21

Break? Isn't that correct?

→ Ответить

2 года назад, # ^ |

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.

→ Ответить

Radewoosh

2 года назад, # ^ |

+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.

→ Ответить

2 года назад, # ^ |

← 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.

→ Ответить

2 года назад, # ^ |

Why the downvotes?

→ Ответить

2 года назад, # |

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 :')

→ Ответить

2 года назад, # ^ |

+16

D accepted in 1 go after the contest :')

→ Ответить

2 года назад, # |

+16

Code of Problem C <<< Solving Problem C

→ Ответить

2 года назад, # |

-13

wait what the heck

Spoiler

→ Ответить

M_bolshakov

2 года назад, # |

-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!

→ Ответить

palindroam

2 года назад, # ^ |

Oh :/

→ Ответить

sstrong

2 года назад, # |

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

→ Ответить

2 года назад, # ^ |

← 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?

→ Ответить

sstrong

2 года назад, # ^ |

← 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.

→ Ответить

Artyom123

2 года назад, # ^ |

+26

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

→ Ответить

2 года назад, # ^ |

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

→ Ответить

sstrong

2 года назад, # ^ |

Oh I see, thanks

→ Ответить

Yahia_Emara

2 года назад, # |

← 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

→ Ответить

2 года назад, # ^ |

-8

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

→ Ответить

dream7

2 года назад, # ^ |

I too solved it using seg tree and DSU.

→ Ответить

Cooper1214

2 года назад, # |

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" :)

→ Ответить

2 года назад, # |

← 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)

→ Ответить

2 года назад, # ^ |

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

→ Ответить

2 года назад, # ^ |

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

→ Ответить

2 года назад, # ^ |

Could you explain the DSU solution?

→ Ответить

2 года назад, # ^ |

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

→ Ответить

rgnerdplayer

2 года назад, # ^ |

-8

171072071

→ Ответить

2 года назад, # ^ |

→ Ответить

2 года назад, # ^ |

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

→ Ответить

2 года назад, # ^ |

Absolutely

→ Ответить

2 года назад, # ^ |

Even this works

→ Ответить

2 года назад, # ^ |

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

→ Ответить

7_7

2 года назад, # |

-10

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

→ Ответить

2 года назад, # |

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

→ Ответить

2 года назад, # |

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

→ Ответить

2 года назад, # ^ |

-10

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

→ Ответить

2 года назад, # ^ |

what is seed?

→ Ответить

2 года назад, # ^ |

← 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.

→ Ответить

2 года назад, # ^ |

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

→ Ответить

2 года назад, # ^ |

-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.

→ Ответить

2 года назад, # ^ |

can you do something ? Newtech66

→ Ответить

MrOtter

2 года назад, # ^ |

+15

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

→ Ответить

2 года назад, # ^ |

-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?

→ Ответить

ryskmax

2 года назад, # |

Спасибо за раунд авторам, очень интересные задачи!

→ Ответить

dorijanlendvaj

2 года назад, # |

← 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.

→ Ответить

2 года назад, # ^ |

-13

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

→ Ответить

toam

2 года назад, # |

+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)

→ Ответить

2 года назад, # ^ |

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

→ Ответить

FynjyBath

2 года назад, # ^ |

← 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)

→ Ответить

Kyou_mo_kawaii

2 года назад, # ^ |

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

→ Ответить

shivam_360

2 года назад, # |

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

→ Ответить

2 года назад, # ^ |

-13

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

→ Ответить

faizanhussain2310

2 года назад, # ^ |

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

→ Ответить

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

2 года назад, # ^ |

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.

→ Ответить

geranazavr555

2 года назад, # |

+10

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

→ Ответить

ganon

2 года назад, # |

← 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?

→ Ответить