F is shit.

2 года назад, # ^ |

+65

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

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gray

2 года назад, # ^ |

← Rev. 2 →

-48

Mihai Eminescu

Luceafărul

A fost odată ca-n povești, A fost ca niciodată. Din rude mari împărătești, O prea frumoasă fată.

Și era una la părinți Și mândră-n toate cele, Cum e Fecioara între sfinți Și luna între stele.

Din umbra falnicelor bolți Ea pasul și-l îndreaptă Lângă fereastră, unde-n colț Luceafărul așteaptă.

Privea în zare cum pe mări Răsare și străluce, Pe mișcătoarele cărări Corăbii negre duce.

Îl vede azi, îl vede mâini, Astfel dorința-i gata; El iar, privind de săptămâni, Îi cade draga fată.

Cum ea pe coate-și răzima Visând ale ei tâmple, De dorul lui și inima Și sufletu-i se împle.

Și cât de viu s-aprinde el În orișicare sară, Spre umbra negrului castel Când ea o să-i apară.

...

Și pas cu pas pe urma ei Alunecă-n odaie, Țesând cu recile-i scântei O mreajă de văpaie.

Și când în pat se-ntinde drept Copila să se culce, I-atinge mâinile pe piept, I-nchide geana dulce;

Și din oglindă luminiș Pe trupu-i se revarsă, Pe ochii mari, bătând închiși Pe fața ei întoarsă.

Ea îl privea cu un surâs, El tremura-n oglindă, Căci o urma adânc în vis De suflet să se prindă.

Iar ea vorbind cu el în somn, Oftând din greu suspină: - O, dulce-al nopții mele domn, De ce nu vii tu? Vină!

Cobori în jos, luceafăr blând, Alunecând pe-o rază, Pătrunde-n casă și în gând Și viața-mi luminează!

El asculta tremurător, Se aprindea mai tare Și s-arunca fulgerător, Se cufunda în mare;

Și apa unde-au fost căzut În cercuri se rotește, Și din adânc necunoscut Un mândru tânăr crește.

Ușor el trece ca pe prag Pe marginea ferestei Și ține-n mână un toiag Încununat cu trestii.

Părea un tânăr voievod Cu păr de aur moale, Un vânăt giulgi se-ncheie nod Pe umerele goale.

Iar umbra feței străvezii E albă ca de ceară - Un mort frumos cu ochii vii Ce scânteie-n afară.

Din sfera mea venii cu greu Ca să-ți urmez chemarea, Iar cerul este tatăl meu Și mumă-mea e marea.

Ca în cămara ta să vin, Să te privesc de-aproape, Am coborât cu-al meu senin Și m-am născut din ape.

O, vin'! odorul meu nespus, Și lumea ta o lasă; Eu sunt luceafărul de sus, Iar tu să-mi fii mireasă.

Colo-n palate de mărgean Te-oi duce veacuri multe, Și toată lumea-n ocean De tine o s-asculte.

O, ești frumos, cum numa-n vis Un înger se arată, Dară pe calea ce-ai deschis N-oi merge niciodată;

Străin la vorbă și la port, Lucești fără de viață, Căci eu sunt vie, tu ești mort, Și ochiul tău mă-ngheață.

...

Trecu o zi, trecură trei Și iarăși, noaptea, vine Luceafărul deasupra ei Cu razele-i senine.

Ea trebui de el în somn Aminte să-și aducă Și dor de-al valurilor domn De inim-o apucă:

Cobori în jos, luceafăr blând, Alunecând pe-o rază, Pătrunde-n casă și în gând Și viața-mi luminează!

Cum el din cer o auzi, Se stinse cu durere, Iar ceru-ncepe a roti În locul unde piere;

În aer rumene văpăi Se-ntind pe lumea-ntreagă, Și din a chaosului văi Un mândru chip se-ncheagă;

Pe negre vițele-i de păr Coroana-i arde pare, Venea plutind în adevăr Scăldat în foc de soare.

Din negru giulgi se desfășor Marmoreele brațe, El vine trist și gânditor Și palid e la față;

Dar ochii mari și minunați Lucesc adânc himeric, Ca două patimi fără saț Și pline de-ntuneric.

Din sfera mea venii cu greu Ca să te-ascult ș-acuma, Și soarele e tatăl meu, Iar noaptea-mi este muma;

O, vin', odorul meu nespus, Și lumea ta o lasă; Eu sunt luceafărul de sus, Iar tu să-mi fii mireasă.

O, vin', în părul tău bălai S-anin cununi de stele, Pe-a mele ceruri să răsai Mai mândră decât ele.

O, ești frumos cum numa-n vis Un demon se arată, Dară pe calea ce-ai deschis N-oi merge niciodată!

Mă dor de crudul tău amor A pieptului meu coarde, Și ochii mari și grei mă dor, Privirea ta mă arde.

Dar cum ai vrea să mă cobor? Au nu-nțelegi tu oare, Cum că eu sunt nemuritor, Și tu ești muritoare?
Nu caut vorbe pe ales, Nici știu cum aș începe - Deși vorbești pe înțeles, Eu nu te pot pricepe;

Dar dacă vrei cu crezământ Să te-ndrăgesc pe tine, Tu te coboară pe pământ, Fii muritor ca mine.

Tu-mi cei chiar nemurirea mea În schimb pe-o sărutare, Dar voi să știi asemenea Cât te iubesc de tare;

Da, mă voi naște din păcat, Primind o altă lege; Cu vecinicia sunt legat, Ci voi să mă dezlege.

Și se tot duce... S-a tot dus. De dragu-unei copile, S-a rupt din locul lui de sus, Pierind mai multe zile.

...

În vremea asta Cătălin, Viclean copil de casă, Ce umple cupele cu vin Mesenilor la masă,

Un paj ce poartă pas cu pas A-mpărătesii rochii, Băiat din flori și de pripas, Dar îndrăzneț cu ochii,

Cu obrăjei ca doi bujori De rumeni, bată-i vina, Se furișează pânditor Privind la Cătălina.

Dar ce frumoasă se făcu Și mândră, arz-o focul; Ei, Cătălin, acu-i acu Ca să-ți încerci norocul.

Și-n treacăt o cuprinse lin Într-un ungher degrabă. - Da' ce vrei, mări Cătălin? Ia du-t' de-ți vezi de treabă.

Ce voi? Aș vrea să nu mai stai Pe gânduri totdeauna, Să râzi mai bine și să-mi dai O gură, numai una.
Dar nici nu știu măcar ce-mi ceri, Dă-mi pace, fugi departe - O, de luceafărul din cer M-a prins un dor de moarte.
Dacă nu știi, ți-aș arăta Din bob în bob amorul, Ci numai nu te mânia, Ci stai cu binișorul.

Cum vânătoru-ntinde-n crâng La păsărele lațul, Când ți-oi întinde brațul stâng Să mă cuprinzi cu brațul;

Și ochii tăi nemișcători Sub ochii mei rămâie... De te înalț de subsuori Te-nalță din călcâie;

Când fața mea se pleacă-n jos, În sus rămâi cu fața, Să ne privim nesățios Și dulce toată viața;

Și ca să-ți fie pe deplin Iubirea cunoscută, Când sărutându-te mă-nclin, Tu iarăși mă sărută.

Ea-l asculta pe copilaș Uimită și distrasă, Și rușinos și drăgălaș, Mai nu vrea, mai se lasă,

Și-i zice-ncet: — Încă de mic Te cunoșteam pe tine, Și guraliv și de nimic, Te-ai potrivi cu mine...

Dar un luceafăr, răsărit Din liniștea uitării, Dă orizon nemărginit Singurătății mării;

Și tainic genele le plec, Căci mi le umple plânsul Când ale apei valuri trec Călătorind spre dânsul;

Lucește c-un amor nespus, Durerea să-mi alunge, Dar se înalță tot mai sus, Ca să nu-l pot ajunge.

Pătrunde trist cu raze reci Din lumea ce-l desparte... În veci îl voi iubi și-n veci Va rămânea departe...

De-aceea zilele îmi sunt Pustii ca niște stepe, Dar nopțile-s de-un farmec sfânt Ce nu-l mai pot pricepe.

Tu ești copilă, asta e... Hai ș-om fugi în lume, Doar ni s-or pierde urmele Și nu ne-or ști de nume,

Căci amândoi vom fi cuminți, Vom fi voioși și teferi, Vei pierde dorul de părinți Și visul de luceferi.

...

Porni luceafărul. Creșteau În cer a lui aripe, Și căi de mii de ani treceau În tot atâtea clipe.

Un cer de stele dedesubt, Deasupra-i cer de stele - Părea un fulger ne'ntrerupt Rătăcitor prin ele.

Și din a chaosului văi, Jur împrejur de sine, Vedea, ca-n ziua cea dentâi, Cum izvorau lumine;

Cum izvorând îl înconjor Ca niște mări, de-a-notul... El zboară, gând purtat de dor, Pân' piere totul, totul;

Căci unde-ajunge nu-i hotar, Nici ochi spre a cunoaște, Și vremea-ncearcă în zadar Din goluri a se naște.

Nu e nimic și totuși e O sete care-l soarbe, E un adânc asemene Uitării celei oarbe.

De greul negrei vecinicii, Părinte, mă dezleagă Și lăudat pe veci să fii Pe-a lumii scară-ntreagă;

O, cere-mi, Doamne, orice preț Dar dă-mi o altă soarte, Căci tu izvor ești de vieți Și dătător de moarte;

Reia-mi al nemuririi nimb Și focul din privire, Și pentru toate dă-mi în schimb O oră de iubire...

Din chaos, Doamne,-am apărut Și m-aș întoarce-n chaos... Și din repaos m-am născut, Mi-e sete de repaos.

Hyperion, ce din genuni Răsai c-o-ntreagă lume, Nu cere semne și minuni Care n-au chip și nume;

Tu vrei un om să te socoți Cu ei să te asameni? Dar piară oamenii cu toți, S-ar naște iarăși oameni.

Ei numai doar durează-n vânt Deșerte idealuri - Când valuri află un mormânt, Răsar în urmă valuri;

Ei doar au stele cu noroc Și prigoniri de soarte, Noi nu avem nici timp, nici loc Și nu cunoaștem moarte.

Din sânul vecinicului ieri Trăiește azi ce moare, Un soare de s-ar stinge-n cer S-aprinde iarăși soare;

Părând pe veci a răsări, Din urmă moartea-l paște, Căci toți se nasc spre a muri Și mor spre a se naște.

Iar tu, Hyperion, rămâi Oriunde ai apune... Cere-mi cuvântul meu dentâi - Să-ți dau înțelepciune?

Vrei să dau glas acelei guri, Ca dup-a ei cântare Să se ia munții cu păduri Și insulele-n mare?

Vrei poate-n faptă să arăți Dreptate și tărie? Ți-aș da pământul în bucăți Să-l faci împărăție.

Îți dau catarg lângă catarg, Oștiri spre a străbate Pământu-n lung și marea-n larg, Dar moartea nu se poate...

Și pentru cine vrei să mori? Întoarce-te, te-ndreaptă Spre-acel pământ rătăcitor Și vezi ce te așteaptă.

...

În locul lui menit din cer Hyperion se-ntoarse Și, ca și-n ziua cea de ieri, Lumina și-o revarsă.

Căci este sara-n asfințit Și noaptea o să-nceapă; Răsare luna liniștit Și tremurând din apă

Și umple cu-ale ei scântei Cărările din crânguri. Sub șirul lung de mândri tei Ședeau doi tineri singuri:

O, lasă-mi capul meu pe sân, Iubito, să se culce Sub raza ochiului senin Și negrăit de dulce;

Cu farmecul luminii reci Gândirile străbate-mi, Revarsă liniște de veci Pe noaptea mea de patimi.

Și de asupra mea rămâi Durerea mea de-o curmă, Căci ești iubirea mea dentâi Și visul meu din urmă.

Hyperion vedea de sus Uimirea-n a lor față: Abia un braț pe gât i-a pus Și ea l-a prins în brațe...

Miroase florile-argintii Și cad, o dulce ploaie, Pe creștetele-a doi copii Cu plete lungi, bălaie.

Ea, îmbătată de amor, Ridică ochii. Vede Luceafărul. Și-ncetișor Dorințele-i încrede:

Cobori în jos, luceafăr blând, Alunecând pe-o rază, Pătrunde-n codru și în gând, Norocu-mi luminează!

El tremură ca alte dăți În codri și pe dealuri, Călăuzind singurătăți De mișcătoare valuri;

Dar nu mai cade ca-n trecut În mări din tot înaltul: - Ce-ți pasă ție, chip de lut, Dac-oi fi eu sau altul?

Trăind în cercul vostru strâmt Norocul vă petrece, Ci eu în lumea mea mă simt Nemuritor și rece.

1883, aprilie

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2 года назад, # ^ |

+12

you've reposted in the wrong connected component

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User_Carrot

2 года назад, # ^ |

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

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Greedious

2 года назад, # ^ |

← Rev. 2 →

+60

posting mind blowing meme to get upvotes :

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shivansh1102

2 года назад, # |

+60

As the leamder

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absolute-mess

2 года назад, # ^ |

+45

Domne

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anubhavdhar

2 года назад, # |

+52

As an author

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Tom66

2 года назад, # |

+30

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

When will scoring distribution come out?

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physics0523

2 года назад, # |

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

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glebustim

2 года назад, # |

+30

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

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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.

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.Miikka.

2 года назад, # |

← Rev. 2 →

+15

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

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2 года назад, # ^ |

+14

As an author, me too.

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.Miikka.

2 года назад, # ^ |

Looking forward to it :p

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2 года назад, # |

-262

As an author

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2 года назад, # ^ |

-8

omg that thumbs up emoji looks fabulous

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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)

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2 года назад, # ^ |

-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 года назад, # ^ |

-251

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

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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...

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flakes24

2 года назад, # ^ |

got downvoted for being right

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Jynxzi

2 года назад, # ^ |

now this is a comeback

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2 года назад, # ^ |

help me, this comeback has reversed somehow

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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!

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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

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AntiXoR

2 года назад, # |

← Rev. 2 →

IIT KGP !orz

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cf-coder

2 года назад, # ^ |

What is IIT KGP?

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drakeramoray27

2 года назад, # ^ |

+10

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

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ZyadFva

2 года назад, # |

As a pupil contestant

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Hamroz_JONN

2 года назад, # ^ |

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

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Echp0chmak

2 года назад, # ^ |

and what's so wrong with it?

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ZyadFva

2 года назад, # ^ |

-8

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

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whynesspower

2 года назад, # ^ |

why is it unrated?

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ZyadFva

2 года назад, # ^ |

one of problems was copied from another judge.

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allal

2 года назад, # |

I hope I can do at least the first one

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Omar_Hafez

2 года назад, # |

+91

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2 года назад, # ^ |

+38

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

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

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2 года назад, # ^ |

first time ?

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tiasmondal

2 года назад, # |

+50

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

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I.am.Moana.Of.Motunui

2 года назад, # |

← 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 года назад, # ^ |

Good luck

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islam231

2 года назад, # |

← Rev. 2 →

Is it rated for DIV 2 ?

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2 года назад, # ^ |

+14

Yes, it is rated for all divisions.

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pirate_hunter_zoro

2 года назад, # ^ |

not any more xD

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Noobish_Monk

2 года назад, # |

Hoping to solve 4 problems, would be great

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2 года назад, # ^ |

-259

Wishing you luck, to solve even more

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nowadays-

2 года назад, # ^ |

Hope you can solve more problems too

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Retired_Snow

2 года назад, # |

Hope to become as soon as possible Candidate Master.

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2 года назад, # ^ |

-266

All the best

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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>

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thelastdinner

2 года назад, # ^ |

+32

However, A new comment will be shown in blue background

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Abito

2 года назад, # ^ |

hidden content*

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Allen123

2 года назад, # |

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

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2 года назад, # ^ |

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 года назад, # ^ |

+23

god, you're everywhere

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Persistence_Phoenix_010

2 года назад, # |

-12

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

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hackerbhaiya

2 года назад, # ^ |

+22

It's IND vs SL!

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mr-strange

2 года назад, # |

All the Best EveryOne!

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nowadays-

2 года назад, # |

Good luck to all

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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?

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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.

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Saucemaster102

2 года назад, # |

+17

C was so good :D

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AlexMari

2 года назад, # |

Very goood problems A, B, C!

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Ayanokouji

2 года назад, # |

How to solve C? Any hints

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OutrunMyGun

2 года назад, # ^ |

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

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2 года назад, # ^ |

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

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2 года назад, # ^ |

I just use DSU.

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2 года назад, # ^ |

← Rev. 2 →

How is its time complexity?

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2 года назад, # ^ |

I don't calculate but get only 31 ms.

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2 года назад, # ^ |

← Rev. 2 →

+22

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

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Ayanokouji

2 года назад, # ^ |

← Rev. 3 →

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

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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.

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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.

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doraty228

2 года назад, # ^ |

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

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hzr

2 года назад, # |

+233

F is shit.

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Lightnessqwq

2 года назад, # ^ |

+73

Can't agree more.

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2 года назад, # |

Problem C was nice.

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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

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2 года назад, # ^ |

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

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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

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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

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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?

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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

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2 года назад, # ^ |

I see, thanks once again :D

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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.

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2 года назад, # ^ |

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

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2 года назад, # ^ |

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

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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.

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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.

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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 ;-;

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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).

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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!!

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mukulrawat_03

2 года назад, # ^ |

i messed up preety badly at c

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2 года назад, # |

+40

Problem F is boring and hard to implement :(

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shauryasinha020

2 года назад, # ^ |

Any hints for D?

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2 года назад, # ^ |

The connected components are trees.

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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 )?

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2 года назад, # ^ |

+30

Random algorithm .

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nidaboge

2 года назад, # ^ |

How to solve d with random algorithm

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

2 года назад, # ^ |

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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]).

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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.

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OutrunMyGun

2 года назад, # ^ |

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

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2 года назад, # ^ |

I mean the connected components formed by same color.

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OutrunMyGun

2 года назад, # ^ |

← Rev. 2 →

Oh, okey when.

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2 года назад, # ^ |

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

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2 года назад, # ^ |

FST now :(

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shauryasinha020

2 года назад, # |

hints for D?

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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

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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

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Wielomian

2 года назад, # ^ |

Hint 1

Hint 2

Hint 3

Solution

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2 года назад, # ^ |

How is this optimal?

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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.

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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.

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dl720125

2 года назад, # ^ |

+10

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

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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.

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__hermit__

2 года назад, # |

What is the expected time complexity for D?

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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.

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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.

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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.

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2 года назад, # ^ |

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

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feecIe6418

2 года назад, # |

+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 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)

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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.

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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.

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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!

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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.

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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

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Nikhil19259

2 года назад, # ^ |

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

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2 года назад, # ^ |

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

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Nikhil19259

2 года назад, # ^ |

Can you share the your original submission link?

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2 года назад, # ^ |

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

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Varad

2 года назад, # ^ |

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

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2 года назад, # ^ |

thanks a lot @varad2002

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Varad

2 года назад, # ^ |

You’re welcome :)

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

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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 .

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savsmail

2 года назад, # ^ |

Wow, that explains a lot. Thank you!

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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++)

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2 года назад, # ^ |

thanks a lot for your suggestions.@Nikhil19259

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dsquare

2 года назад, # |

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

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Rikitexh

2 года назад, # |

hint for c?

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Citypop

2 года назад, # ^ |

)(

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2 года назад, # ^ |

+12

Spoiler 1

Spoiler 2

Solution

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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

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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!

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Theo830

2 года назад, # |

Any hints for E?

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bajablast

2 года назад, # ^ |

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

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TanSinTheta

2 года назад, # |

Really nice problems ! Keep it up authors ^_^

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rgnerdplayer

2 года назад, # |

+16

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

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2 года назад, # ^ |

+20

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

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rgnerdplayer

2 года назад, # ^ |

+13

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

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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...

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Radewoosh

2 года назад, # ^ |

+21

Break? Isn't that correct?

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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.

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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.

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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.

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2 года назад, # ^ |

Why the downvotes?

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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 :')

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2 года назад, # ^ |

+16

D accepted in 1 go after the contest :')

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2 года назад, # |

+16

Code of Problem C <<< Solving Problem C

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2 года назад, # |

-13

wait what the heck

Spoiler

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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!

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palindroam

2 года назад, # ^ |

Oh :/

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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.

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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?

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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.

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Artyom123

2 года назад, # ^ |

+26

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

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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

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sstrong

2 года назад, # ^ |

Oh I see, thanks

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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.me/contest/1726/submission/171128759

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2 года назад, # ^ |

-8

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

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dream7

2 года назад, # ^ |

I too solved it using seg tree and DSU.

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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" :)

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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)

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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.

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2 года назад, # ^ |

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

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2 года назад, # ^ |

Could you explain the DSU solution?

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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

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rgnerdplayer

2 года назад, # ^ |

-8

171072071

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2 года назад, # ^ |

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2 года назад, # ^ |

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

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2 года назад, # ^ |

Absolutely

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2 года назад, # ^ |

Even this works

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2 года назад, # ^ |

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

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7_7

2 года назад, # |

-10

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

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2 года назад, # |

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 года назад, # |

My exactly same code got ac after system test

AC code : https://codeforces.me/contest/1726/submission/171155848 TLE code : https://codeforces.me/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 года назад, # ^ |

-10

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

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2 года назад, # ^ |

what is seed?

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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.

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2 года назад, # ^ |

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

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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.

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2 года назад, # ^ |

can you do something ? Newtech66

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MrOtter

2 года назад, # ^ |

+15

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

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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?

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ryskmax

2 года назад, # |

Спасибо за раунд авторам, очень интересные задачи!

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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.

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2 года назад, # ^ |

-13

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

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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)

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2 года назад, # ^ |

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 года назад, # ^ |

← Rev. 2 →

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

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Kyou_mo_kawaii

2 года назад, # ^ |

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

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shivam_360

2 года назад, # |

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

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2 года назад, # ^ |

-13

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

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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

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

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geranazavr555

2 года назад, # |

+10

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

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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?

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