F is shit.

2 years ago, # ^ |

+65

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

→ Reply

bycicle

2 years ago, # ^ |

-48

As a participant, I will have participated.

Spoiler for upvotes:

→ Reply

2 years ago, # ^ |

+12

you've reposted in the wrong connected component

→ Reply

User_Carrot

2 years ago, # ^ |

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

→ Reply

Greedious

2 years ago, # ^ |

← Rev. 2 →

+60

posting mind blowing meme to get upvotes :

→ Reply

shivansh1102

2 years ago, # |

+60

As the leamder

→ Reply

absolute-mess

2 years ago, # ^ |

+45

Domne

→ Reply

anubhavdhar

2 years ago, # |

+52

As an author

→ Reply

Tom66

2 years ago, # |

+30

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

When will scoring distribution come out?

→ Reply

physics0523

2 years ago, # |

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

→ Reply

glebustim

2 years ago, # |

+30

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

→ Reply

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.

→ Reply

.Miikka.

2 years ago, # |

← Rev. 2 →

+15

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

→ Reply

2 years ago, # ^ |

+14

As an author, me too.

→ Reply

.Miikka.

2 years ago, # ^ |

Looking forward to it :p

→ Reply

2 years ago, # |

-262

As an author

→ Reply

2 years ago, # ^ |

-8

omg that thumbs up emoji looks fabulous

→ Reply

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)

→ Reply

2 years ago, # ^ |

-22

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

→ Reply

2 years ago, # ^ |

-251

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

→ Reply

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...

→ Reply

flakes24

2 years ago, # ^ |

got downvoted for being right

→ Reply

Jynxzi

2 years ago, # ^ |

now this is a comeback

→ Reply

2 years ago, # ^ |

help me, this comeback has reversed somehow

→ Reply

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!

→ Reply

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

→ Reply

AntiXoR

2 years ago, # |

← Rev. 2 →

IIT KGP !orz

→ Reply

cf-coder

2 years ago, # ^ |

What is IIT KGP?

→ Reply

drakeramoray27

2 years ago, # ^ |

+10

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

→ Reply

ZyadFva

2 years ago, # |

As a pupil contestant

→ Reply

Hamroz_JONN

2 years ago, # ^ |

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

→ Reply

Echp0chmak

2 years ago, # ^ |

and what's so wrong with it?

→ Reply

ZyadFva

2 years ago, # ^ |

-8

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

→ Reply

whynesspower

2 years ago, # ^ |

why is it unrated?

→ Reply

ZyadFva

2 years ago, # ^ |

one of problems was copied from another judge.

→ Reply

allal

2 years ago, # |

I hope I can do at least the first one

→ Reply

Omar_Hafez

2 years ago, # |

+91

→ Reply

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

→ Reply

2 years ago, # ^ |

first time ?

→ Reply

tiasmondal

2 years ago, # |

+50

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

→ Reply

SalmaHatem_

2 years ago, # |

← Rev. 2 →

+10

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

→ Reply

nowadays-

2 years ago, # ^ |

Good luck

→ Reply

islam231

2 years ago, # |

← Rev. 2 →

Is it rated for DIV 2 ?

→ Reply

2 years ago, # ^ |

+14

Yes, it is rated for all divisions.

→ Reply

pirate_hunter_zoro

2 years ago, # ^ |

not any more xD

→ Reply

Noobish_Monk

2 years ago, # |

Hoping to solve 4 problems, would be great

→ Reply

2 years ago, # ^ |

-259

Wishing you luck, to solve even more

→ Reply

nowadays-

2 years ago, # ^ |

Hope you can solve more problems too

→ Reply

Retired_Snow

2 years ago, # |

Hope to become as soon as possible Candidate Master.

→ Reply

2 years ago, # ^ |

-266

All the best

→ Reply

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>

→ Reply

thelastdinner

2 years ago, # ^ |

+32

However, A new comment will be shown in blue background

→ Reply

Abito

2 years ago, # ^ |

hidden content*

→ Reply

Allen123

2 years ago, # |

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

→ Reply

2 years ago, # ^ |

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

→ Reply

pigmike

2 years ago, # ^ |

+23

god, you're everywhere

→ Reply

Silly__coder__

2 years ago, # |

-12

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

→ Reply

hackerbhaiya

2 years ago, # ^ |

+22

It's IND vs SL!

→ Reply

mr-strange

2 years ago, # |

All the Best EveryOne!

→ Reply

nowadays-

2 years ago, # |

Good luck to all

→ Reply

Ahmed_AboRadwan

2 years ago, # |

Sometimes you win, sometimes you learn

→ Reply

2 years ago, # |

We just got trolled

→ Reply

2 years ago, # |

How to solve D?

→ Reply

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.

→ Reply

Saucemaster102

2 years ago, # |

+17

C was so good :D

→ Reply

AlexMari

2 years ago, # |

Very goood problems A, B, C!

→ Reply

Ayanokouji

2 years ago, # |

How to solve C? Any hints

→ Reply

OutrunMyGun

2 years ago, # ^ |

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

→ Reply

2 years ago, # ^ |

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

→ Reply

2 years ago, # ^ |

I just use DSU.

→ Reply

2 years ago, # ^ |

← Rev. 2 →

How is its time complexity?

→ Reply

2 years ago, # ^ |

I don't calculate but get only 31 ms.

→ Reply

2 years ago, # ^ |

← Rev. 2 →

+22

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

→ Reply

Ayanokouji

2 years ago, # ^ |

← Rev. 3 →

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

→ Reply

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.

→ Reply

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.

→ Reply

doraty228

2 years ago, # ^ |

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

→ Reply

hzr

2 years ago, # |

+233

F is shit.

→ Reply

Lightnessqwq

2 years ago, # ^ |

+73

Can't agree more.

→ Reply

2 years ago, # |

Problem C was nice.

→ Reply

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

→ Reply

2 years ago, # ^ |

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

→ Reply

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

→ Reply

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

→ Reply

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?

→ Reply

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

→ Reply

2 years ago, # ^ |

I see, thanks once again :D

→ Reply

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.

→ Reply

2 years ago, # ^ |

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

→ Reply

2 years ago, # ^ |

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

→ Reply

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.

→ Reply

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.

→ Reply

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 ;-;

→ Reply

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).

→ Reply

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!!

→ Reply

mukulrawat_03

2 years ago, # ^ |

i messed up preety badly at c

→ Reply

2 years ago, # |

+40

Problem F is boring and hard to implement :(

→ Reply

shauryasinha020

2 years ago, # ^ |

Any hints for D?

→ Reply

2 years ago, # ^ |

The connected components are trees.

→ Reply

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 )?

→ Reply

2 years ago, # ^ |

+30

Random algorithm .

→ Reply

nidaboge

2 years ago, # ^ |

How to solve d with random algorithm

→ Reply

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

2 years ago, # ^ |

→ Reply

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]).

→ Reply

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.

→ Reply

OutrunMyGun

2 years ago, # ^ |

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

→ Reply

2 years ago, # ^ |

I mean the connected components formed by same color.

→ Reply

OutrunMyGun

2 years ago, # ^ |

← Rev. 2 →

Oh, okey when.

→ Reply

2 years ago, # ^ |

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

→ Reply

2 years ago, # ^ |

FST now :(

→ Reply

shauryasinha020

2 years ago, # |

hints for D?

→ Reply

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

→ Reply

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

→ Reply

Wielomian

2 years ago, # ^ |

Hint 1

Hint 2

Hint 3

Solution

→ Reply

2 years ago, # ^ |

How is this optimal?

→ Reply

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.

→ Reply

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.

→ Reply

dl720125

2 years ago, # ^ |

+10

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

→ Reply

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.

→ Reply

__hermit__

2 years ago, # |

What is the expected time complexity for D?

→ Reply

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.

→ Reply

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.

→ Reply

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.

→ Reply

2 years ago, # ^ |

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

→ Reply

feecIe6418

2 years ago, # |

+41

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

→ Reply

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)

→ Reply

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.

→ Reply

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.

→ Reply

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!

→ Reply

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.

→ Reply

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

→ Reply

Nikhil19259

2 years ago, # ^ |

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

→ Reply

2 years ago, # ^ |

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

→ Reply

Nikhil19259

2 years ago, # ^ |

Can you share the your original submission link?

→ Reply

2 years ago, # ^ |

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

→ Reply

Varad

2 years ago, # ^ |

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

→ Reply

2 years ago, # ^ |

thanks a lot @varad2002

→ Reply

Varad

2 years ago, # ^ |

You’re welcome :)

→ Reply

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

→ Reply

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 .

→ Reply

savsmail

2 years ago, # ^ |

Wow, that explains a lot. Thank you!

→ Reply

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++)

→ Reply

2 years ago, # ^ |

thanks a lot for your suggestions.@Nikhil19259

→ Reply

d_sqaure

2 years ago, # |

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

→ Reply

Rikitexh

2 years ago, # |

hint for c?

→ Reply

Citypop

2 years ago, # ^ |

)(

→ Reply

2 years ago, # ^ |

+12

Spoiler 1

Spoiler 2

Solution

→ Reply

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!

→ Reply

Theo830

2 years ago, # |

Any hints for E?

→ Reply

bajablast

2 years ago, # ^ |

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

→ Reply

TanSinTheta

2 years ago, # |

Really nice problems ! Keep it up authors ^_^

→ Reply

rgnerdplayer

2 years ago, # |

+16

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

→ Reply

2 years ago, # ^ |

+20

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

→ Reply

rgnerdplayer

2 years ago, # ^ |

+13

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

→ Reply

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...

→ Reply

Radewoosh

2 years ago, # ^ |

+21

Break? Isn't that correct?

→ Reply

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.

→ Reply

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.

→ Reply

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.

→ Reply

2 years ago, # ^ |

Why the downvotes?

→ Reply

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 :')

→ Reply

2 years ago, # ^ |

+16

D accepted in 1 go after the contest :')

→ Reply

2 years ago, # |

+16

Code of Problem C <<< Solving Problem C

→ Reply

2 years ago, # |

-13

wait what the heck

Spoiler

→ Reply

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!

→ Reply

palindroam

2 years ago, # ^ |

Oh :/

→ Reply

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.

→ Reply

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?

→ Reply

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.

→ Reply

Artyom123

2 years ago, # ^ |

+26

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

→ Reply

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

→ Reply

sstrong

2 years ago, # ^ |

Oh I see, thanks

→ Reply

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

→ Reply

2 years ago, # ^ |

-8

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

→ Reply

dream7

2 years ago, # ^ |

I too solved it using seg tree and DSU.

→ Reply

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" :)

→ Reply

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)

→ Reply

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.

→ Reply

2 years ago, # ^ |

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

→ Reply

2 years ago, # ^ |

Could you explain the DSU solution?

→ Reply

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

→ Reply

rgnerdplayer

2 years ago, # ^ |

-8

171072071

→ Reply

2 years ago, # ^ |

→ Reply

2 years ago, # ^ |

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

→ Reply

2 years ago, # ^ |

Absolutely

→ Reply

2 years ago, # ^ |

Even this works

→ Reply

2 years ago, # ^ |

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

→ Reply

7_7

2 years ago, # |

-10

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

→ Reply

2 years ago, # |

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

→ Reply

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

→ Reply

2 years ago, # ^ |

-10

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

→ Reply

2 years ago, # ^ |

what is seed?

→ Reply

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.

→ Reply

2 years ago, # ^ |

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

→ Reply

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.

→ Reply

2 years ago, # ^ |

can you do something ? Newtech66

→ Reply

MrOtter

2 years ago, # ^ |

+15

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

→ Reply

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?

→ Reply

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.

→ Reply

2 years ago, # ^ |

-13

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

→ Reply

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)

→ Reply

2 years ago, # ^ |

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

→ Reply

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)

→ Reply

Kyou_mo_kawaii

2 years ago, # ^ |

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

→ Reply

shivam_360

2 years ago, # |

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

→ Reply

2 years ago, # ^ |

-13

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

→ Reply

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

→ Reply

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.

→ Reply

geranazavr555

2 years ago, # |

+10

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

→ Reply

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?

→ Reply