sshwyR's blog

Codeforces Round #664
By sshwyR, history, 4 years ago, In English

Hello Codeforces! We (dqa2021, Xiejiadong, Retired_xryjr233, Z18 and me) are excited to invite you to take part in Codeforces Round 664 (Div. 1) and Codeforces Round 664 (Div. 2), which will happen on Aug/12/2020 17:35 (Moscow time).

Huge thanks to:

There will be 5 problems in Div.1 round and 6 problems in Div.2 round. You'll be given 2 hours to solve them.

The story of this round is about that man. Instead of displaying his name, I prefer telling one of his legends (or joke):

"I have a 'friend', who makes lots of money every day, earning a billion in the blink of eyes. With a wave of his hand, OIers all over the world will follow him."

You can post your guesses in the comments.

UPD: Score Distribution:

  • Div.1: 500 — 1000 — 1500 — 1750 — 2500
  • Div.2: 500 — 750 — 1000 — 1250 — 1750 — 2250

Good luck!

UPD: Congratulations to the winners!

Div.1:

  1. Benq
  2. ecnerwala
  3. nick452
  4. 244mhq
  5. neal

Div.2:

  1. C.S.T.T
  2. MyLoveKUN
  3. Rchen3
  4. Rainbow_qwq
  5. evilbuggy
  • Vote: I like it
  • +583
  • Vote: I do not like it

Comments Comments (434)
| Write comment?
»
4 years ago, # |
  Vote: I like it +12 Vote: I do not like it

Auto comment: topic has been updated by sshwyR (previous revision, new revision, compare).

→ Reply
  • »
    »
    4 years ago, # ^ |
      Vote: I like it +60 Vote: I do not like it

    What's the point of declaring top 5 people in Div.2, after all they are just Div.1 guys who registered with their alternative account, isn't this unfair? for example just look at the person who is placed 3rd in Div.2, this guy was newbie, like seriously? This disrupts entire rating distribution.

    → Reply
»
4 years ago, # |
← Rev. 4   Vote: I like it +410 Vote: I do not like it

I am tester, please upvote for me! :)

( I'M CUTE, GIVE ME CONTRIBUTION )

→ Reply
»
4 years ago, # |
  Vote: I like it +9 Vote: I do not like it

Is the man is Bill Gates? :D

→ Reply
»
4 years ago, # |
  Vote: I like it 0 Vote: I do not like it

is that man Jack Ma? just a random guess

→ Reply
»
4 years ago, # |
  Vote: I like it +149 Vote: I do not like it

We don't want stories please, as they come in the way of easily comprehending the problem statement. :(

→ Reply
  • »
    »
    4 years ago, # ^ |
      Vote: I like it +77 Vote: I do not like it

    Oh don't worry. The statement is not long and we just use the main character to describe it. The story itself is easy to comprehend :)

    → Reply
    • »
      »
      »
      4 years ago, # ^ |
      ← Rev. 3   Vote: I like it 0 Vote: I do not like it

      For me who is afraid of number theory, this is good news to be happy about. If there is a math problem, I am also ready to change to the alternate.

      → Reply
»
4 years ago, # |
  Vote: I like it +280 Vote: I do not like it

Tester — Planning to participate in the next div 1.
sshwyR — You already tested it.
Tester -

Spoiler
→ Reply
  • »
    »
    4 years ago, # ^ |
      Vote: I like it +65 Vote: I do not like it

    MeToo

    → Reply
  • »
    »
    4 years ago, # ^ |
      Vote: I like it +78 Vote: I do not like it

    This contest was put on hold for a long time due to the difficult Di2A.

    → Reply
    • »
      »
      »
      4 years ago, # ^ |
        Vote: I like it +70 Vote: I do not like it

      difficult Di2A

      Thank you , now excuse me while I go to sleep for 35 hours real quick

      → Reply
      • »
        »
        »
        »
        4 years ago, # ^ |
          Vote: I like it +10 Vote: I do not like it

        Go for 40 , than that would be sacrilegious

        → Reply
        • »
          »
          »
          »
          »
          4 years ago, # ^ |
            Vote: I like it +22 Vote: I do not like it

          Two set violin = small pp

          → Reply
          • »
            »
            »
            »
            »
            »
            4 years ago, # ^ |
              Vote: I like it +4 Vote: I do not like it

            Bass = pp large

            → Reply
            • »
              »
              »
              »
              »
              »
              »
              4 years ago, # ^ |
                Vote: I like it +2 Vote: I do not like it

              What? You bass gang, had some arguments before, but now is just butthurt, VIOLIN > bass. TwoSet responded to all Davie's videos, but Davie just ignored the "bass diss track" (which was asking for a livestream at the end) and the "davie fakes playing the violin". Brett played bass poorly, but it was REAL, but Davie, on the other hand, FAKED playing it. So what side is winning, huh?

              → Reply
    • »
      »
      »
      4 years ago, # ^ |
        Vote: I like it +1 Vote: I do not like it

      Oops, difficult Di2A. So scary....

      → Reply
    • »
      »
      »
      4 years ago, # ^ |
      ← Rev. 2   Vote: I like it 0 Vote: I do not like it

      Is this difficult div2A? Then what about the contest, which was for Russian elementary school students, with a 1700 rating problem A. (it was like 10 contests before) :D

      → Reply
  • »
    »
    4 years ago, # ^ |
    ← Rev. 2   Vote: I like it 0 Vote: I do not like it

    Feeling sad for such testers but they will get contribution.

    → Reply
    • »
      »
      »
      4 years ago, # ^ |
      ← Rev. 2   Vote: I like it 0 Vote: I do not like it

      Well I knew I wont participate in one div1 when I was testing. Anyways I did participated but in unrated mode. Just that I didnt knew I wont be participating on 12 Aug.

      Spoiler
      → Reply
»
4 years ago, # |
  Vote: I like it +60 Vote: I do not like it

Another Chinese Round! This round has special and interesting stories.Good luck and have fun for every contestants. Some information in advance:Those stories may all be about one mysterious person.

[O)]-[(O]/

→ Reply
»
4 years ago, # |
  Vote: I like it +7 Vote: I do not like it

Is the man @MikeMirzayanov ? <3 i have not any guess about his income but i love him and want to see his name in the problem statement ..

→ Reply
»
4 years ago, # |
  Vote: I like it +9 Vote: I do not like it

Mukesh Ambani???

→ Reply
»
4 years ago, # |
  Vote: I like it +32 Vote: I do not like it

Boss Cai!!

→ Reply
»
4 years ago, # |
  Vote: I like it +15 Vote: I do not like it

I am almost sure this man is Jack Ma.

→ Reply
»
4 years ago, # |
  Vote: I like it 0 Vote: I do not like it

I want to know what is Polygon platform ? Strange word for me

→ Reply
  • »
    »
    4 years ago, # ^ |
      Vote: I like it +4 Vote: I do not like it

    A platform for making problems

    https://polygon.codeforces.com/

    → Reply
  • »
    »
    4 years ago, # ^ |
      Vote: I like it +1 Vote: I do not like it

    Polygon in basically the platform for creation of programming contest problems. It also supports problem statement writing, test data preparing, model solutions, judging and automatic validation.

    → Reply
»
4 years ago, # |
  Vote: I like it 0 Vote: I do not like it

I know! The man is Cai Rui[user:Boboniu]!

→ Reply
»
4 years ago, # |
← Rev. 2   Vote: I like it +7 Vote: I do not like it

Intersting fact: tourist will set a new best rating ever on codeforces if he wins this round!

→ Reply
»
4 years ago, # |
  Vote: I like it -18 Vote: I do not like it

Chinese Round?

→ Reply
»
4 years ago, # |
← Rev. 2   Vote: I like it +199 Vote: I do not like it

Fun fact: seeing MiFaFaOvO can't participate in this round, if he doesn't in the next global round, he will vanish from the live standings xD

And "poof": he is gone xD

→ Reply
  • »
    »
    4 years ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    Excuse my ignorance, but what is the rule exactly you're talking about?

    → Reply
    • »
      »
      »
      4 years ago, # ^ |
      ← Rev. 2   Vote: I like it +52 Vote: I do not like it

      Only the users which participated in a rated round within $$$6$$$ months will be included in the Rating Standing.

      The last rated round for him took place on $$$Feb^{17th}$$$.

      → Reply
»
4 years ago, # |
  Vote: I like it +70 Vote: I do not like it

I guess it's CCF

→ Reply
»
4 years ago, # |
  Vote: I like it 0 Vote: I do not like it

I do CP for fun.. What about you guys?

→ Reply
»
4 years ago, # |
  Vote: I like it -118 Vote: I do not like it

BINOD !

→ Reply
  • »
    »
    4 years ago, # ^ |
      Vote: I like it +46 Vote: I do not like it

    Please don't spam Codeforces at least. There are many who don't know the context and it gets annoying after a while.

    → Reply
»
4 years ago, # |
  Vote: I like it 0 Vote: I do not like it

It's Mr-Cai!!! Hahahaha

→ Reply
»
4 years ago, # |
  Vote: I like it 0 Vote: I do not like it

When I see Chinese round, I think of Mathforces.

→ Reply
»
4 years ago, # |
← Rev. 2   Vote: I like it 0 Vote: I do not like it

so... Yet another maximizing profit problems ... or Yet another guessing name problems :v

→ Reply
»
4 years ago, # |
  Vote: I like it 0 Vote: I do not like it

Ronan Ryan?

→ Reply
»
4 years ago, # |
  Vote: I like it 0 Vote: I do not like it

interestingLSY Time for your picture about boboniu!

→ Reply
»
4 years ago, # |
← Rev. 3   Vote: I like it -14 Vote: I do not like it

Ah-oh, jqdai0815 as a tester, which means we can't see his competition with tourist this time. A huge pity!

→ Reply
  • »
    »
    4 years ago, # ^ |
      Vote: I like it -71 Vote: I do not like it

    Either way ,he will never compete against tourist ,after the time he gained 1st spot . He just prays for some bad luck for tourist ,so that he can remain on top.

    Before that ,i Never thought ,such legendary grandmaster would ever fear to directly compete against someone. But anyway that someone tourist is not ordinary guy.

    → Reply
  • »
    »
    4 years ago, # ^ |
      Vote: I like it +49 Vote: I do not like it

    But It's still an interesting and exciting contest

    → Reply
»
4 years ago, # |
  Vote: I like it +69 Vote: I do not like it

He must be Zide Du(杜子德) from the CCF.

→ Reply
  • »
    »
    4 years ago, # ^ |
      Vote: I like it +38 Vote: I do not like it

    China Computer Federation or China Collecting-money Federation?

    → Reply
»
4 years ago, # |
  Vote: I like it 0 Vote: I do not like it

Is him, right ?

→ Reply
»
4 years ago, # |
  Vote: I like it -7 Vote: I do not like it

Please provide short statements for problems and understandable english,I am learning english .Thanks

→ Reply
  • »
    »
    4 years ago, # ^ |
      Vote: I like it +2 Vote: I do not like it

    Copy to baidu translate (

    Website: fanyi.baidu.com

    → Reply
»
4 years ago, # |
← Rev. 11   Vote: I like it -64 Vote: I do not like it

[Deleted]

→ Reply
  • »
    »
    4 years ago, # ^ |
      Vote: I like it +39 Vote: I do not like it

    You take slander as freedom?

    → Reply
  • »
    »
    4 years ago, # ^ |
      Vote: I like it +30 Vote: I do not like it

    You should know it's not polite to comment others like that

    → Reply
  • »
    »
    4 years ago, # ^ |
      Vote: I like it +37 Vote: I do not like it

    Freedom is not your excuse for slandering.

    → Reply
  • »
    »
    4 years ago, # ^ |
      Vote: I like it +29 Vote: I do not like it

    Although it is your right to express your own ideas,but your last comment was unsuitable indeed in some way.And codeforces delete your comment for a great many people downvote you.I think it is reasonable.

    → Reply
»
4 years ago, # |
  Vote: I like it +49 Vote: I do not like it

See this 89567737

→ Reply
»
4 years ago, # |
  Vote: I like it -6 Vote: I do not like it

What's happening to "timeanddate.com"? :|

→ Reply
  • »
    »
    4 years ago, # ^ |
    ← Rev. 2   Vote: I like it +11 Vote: I do not like it

    The website works well for me, probably it's because of your network problem.

    And I don't think these sorts of comments are meaningful here

    → Reply
»
4 years ago, # |
  Vote: I like it +24 Vote: I do not like it

Boss Cai -- boboniu

→ Reply
»
4 years ago, # |
← Rev. 2   Vote: I like it +6 Vote: I do not like it

Makes lots of money everyday. That must be dzd from CCF(China Collecting-money Foundation)!

→ Reply
»
4 years ago, # |
  Vote: I like it 0 Vote: I do not like it

I became psychopath because of the little pony, please make sure that the problems have short statements. with appreciation to your great story.

→ Reply
»
4 years ago, # |
  Vote: I like it 0 Vote: I do not like it

I really hope that the problems are not as hard as the quiz.....

→ Reply
»
4 years ago, # |
  Vote: I like it +11 Vote: I do not like it

How many number of common problems ?

→ Reply
»
4 years ago, # |
  Vote: I like it 0 Vote: I do not like it

Richie Rich?

→ Reply
»
4 years ago, # |
  Vote: I like it +10 Vote: I do not like it

It could be suneo's dad from doraemon xD

→ Reply
»
4 years ago, # |
  Vote: I like it 0 Vote: I do not like it

is it jeff bezos ?

→ Reply
»
4 years ago, # |
  Vote: I like it +7 Vote: I do not like it

Olers?

→ Reply
»
4 years ago, # |
  Vote: I like it +1 Vote: I do not like it

Is that man Mark Zuckerberg??

→ Reply
»
4 years ago, # |
← Rev. 2   Vote: I like it +13 Vote: I do not like it
The Man is:
→ Reply
»
4 years ago, # |
  Vote: I like it +9 Vote: I do not like it

Is the man Mike Mirzayanov ?

→ Reply
»
4 years ago, # |
  Vote: I like it 0 Vote: I do not like it

JPow?

→ Reply
»
4 years ago, # |
  Vote: I like it +41 Vote: I do not like it

If there are unnecessary stories, please, do mark them in Italic or any other way. I've faced a lot of issues about bad storytelling in Codeforces. :)

→ Reply
»
4 years ago, # |
  Vote: I like it +3 Vote: I do not like it

What are "Olers"?

→ Reply
  • »
    »
    4 years ago, # ^ |
      Vote: I like it +14 Vote: I do not like it

    The people like us. Take part in OI. OI, Olympics Information. OIer, the people learn OI. OIers is plural form of OIer. Understand? :)

    I'm not good at English. I hope you can understand xD

    → Reply
  • »
    »
    4 years ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    OIers are coders who participate in programming competitions (mostly used in China I guess). OI = Olympiad Informatics

    → Reply
  • »
    »
    4 years ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    It's what Chinese called people who participate in Olympic in Informatics contests.

    → Reply
  • »
    »
    4 years ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    The goalkeeper of RCD Espanyol, click the link for more details.

    https://www.transfermarkt.de/oier-olazabal/profil/spieler/8176

    Since Wu Lei (the best football player in China) is also in this team, we use the word "Oier" to show our respect to RCD Espanyol so that they pay Wu Lei on time.

    → Reply
»
4 years ago, # |
  Vote: I like it -14 Vote: I do not like it

i guess Jack Ma .

→ Reply
»
4 years ago, # |
  Vote: I like it +5 Vote: I do not like it

I guess he's kkksc03, head of Luogu(an oj in China). For every monthly contest in Luogu, if you want to watch the video tutorial, you must pay 10￥ first. And what's more there's lots of OI online lesson made by Luogu in every summer and they're not free! So he can make lots of money in this way.

→ Reply
  • »
    »
    4 years ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    Actually, according to himself, the money is not paid to him directly and a lot of money is spent in maintaining Luogu's website. And the lessons' money will be given to the teachers. So I'm very sure that it's not him.

    → Reply
»
4 years ago, # |
  Vote: I like it +14 Vote: I do not like it

Any hint about no of common problems or score distribution ?

→ Reply
»
4 years ago, # |
  Vote: I like it +19 Vote: I do not like it

Score Distribution?

→ Reply
»
4 years ago, # |
  Vote: I like it 0 Vote: I do not like it

Obviously,the man is DUZIDE:P

→ Reply
»
4 years ago, # |
  Vote: I like it +14 Vote: I do not like it

Score Distribution?

→ Reply
  • »
    »
    4 years ago, # ^ |
      Vote: I like it +8 Vote: I do not like it

    Sorry for the late publishment of score distribution :)

    → Reply
»
4 years ago, # |
← Rev. 2   Vote: I like it -9 Vote: I do not like it

I have a request for MikeMirzayanov.. can we have TEAM contest at codeforces.

→ Reply
»
4 years ago, # |
  Vote: I like it +4 Vote: I do not like it

It's €€£

→ Reply
»
4 years ago, # |
  Vote: I like it +7 Vote: I do not like it

sshwyR how many problems will be common in both divisions?

→ Reply
»
4 years ago, # |
  Vote: I like it +4 Vote: I do not like it

Auto comment: topic has been updated by sshwyR (previous revision, new revision, compare).

→ Reply
»
4 years ago, # |
  Vote: I like it +36 Vote: I do not like it

Today's tester busy with increasing their contribution .

→ Reply
»
4 years ago, # |
  Vote: I like it +13 Vote: I do not like it

Div2 Problem E with 1750 (^_^)...it should be solvable

→ Reply
»
4 years ago, # |
  Vote: I like it +6 Vote: I do not like it

iam not a tester, but please upvote for my pp :).......

→ Reply
»
4 years ago, # |
  Vote: I like it 0 Vote: I do not like it

The man must be, President of IOI: Richard Forster

→ Reply
»
4 years ago, # |
  Vote: I like it +8 Vote: I do not like it

Zhengru IOI will be internationalized.

→ Reply
»
4 years ago, # |
  Vote: I like it 0 Vote: I do not like it

Karuna! :D

→ Reply
»
4 years ago, # |
  Vote: I like it 0 Vote: I do not like it

An off-topic question: how can I view the past contests right before contests?

→ Reply
»
4 years ago, # |
  Vote: I like it +8 Vote: I do not like it

its boboniu

→ Reply
»
4 years ago, # |
  Vote: I like it +1 Vote: I do not like it

Too much if else... :(

→ Reply
»
4 years ago, # |
  Vote: I like it 0 Vote: I do not like it

Normal Div1:

Try to solve more problems

Horrible Round #664 Div1:

Try to solve the first problem faster

→ Reply
  • »
    »
    4 years ago, # ^ |
      Vote: I like it +27 Vote: I do not like it

    No,you should try to avoid FST instead of solving problems faster. It's my final standing: xjpbmbrm.png

    (I was rk128 before the system test...)

    → Reply
»
4 years ago, # |
  Vote: I like it 0 Vote: I do not like it

If I accept no question am i still rated?

→ Reply
  • »
    »
    4 years ago, # ^ |
      Vote: I like it +28 Vote: I do not like it 
    if(submit any problem solution) Yes
else No
accept or not doesn't matter
    → Reply
»
4 years ago, # |
  Vote: I like it +10 Vote: I do not like it

why they set so hard

→ Reply
»
4 years ago, # |
  Vote: I like it +12 Vote: I do not like it

Was this the "Single Test Case" Contest?

→ Reply
»
4 years ago, # |
  Vote: I like it -7 Vote: I do not like it

Really nice problems and clear statements. Enjoyed taking this contest :D

→ Reply
»
4 years ago, # |
  Vote: I like it -8 Vote: I do not like it

Codeforces rounds are great, they teach us how to look at a problem in a simple manner, without overcomplicating it. orz Great contest.

→ Reply
»
4 years ago, # |
  Vote: I like it +9 Vote: I do not like it

Well well well, once again someone posted solutions on youtube during the contest, you can even see his username in the video, i wonder if these guys are shameless or what Video link Username: Khayrulmithu

→ Reply
  • »
    »
    4 years ago, # ^ |
      Vote: I like it +4 Vote: I do not like it

    I don't know why but that video sounds exactly like an airline pilot's announcement.

    → Reply
  • »
    »
    4 years ago, # ^ |
      Vote: I like it +5 Vote: I do not like it

    And he got it wrong on test 10 :D

    → Reply
»
4 years ago, # |
  Vote: I like it -6 Vote: I do not like it

I really liked d1 B, but...

Starting from any vertex u

This would be much more understandable if it said from all vertices. Wasted lots of time solving the wrong problem.

→ Reply
  • »
    »
    4 years ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    That could confuse people into thinking that it's possible to start from multiple vertices simultaneously, I think the clearest is "for all vertices u, starting from u, ..."

    → Reply
  • »
    »
    4 years ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    I'm pretty sure that "any" was actually used correctly here

    → Reply
»
4 years ago, # |
  Vote: I like it -14 Vote: I do not like it

What was the point of swapping X and Y in Div2B?

→ Reply
»
4 years ago, # |
  Vote: I like it +10 Vote: I do not like it

Does this works for Div-2 E — Make a graph $$$g[i][j][k]$$$ = how many edges exist such that its head is on node with $$$i$$$ outgoing edges and tail is in node with $$$j$$$ outgoing edges and it is $$$k'th$$$ smallest one. Then we brute force every $$$c_i$$$. That will be $$$O(k!k^2)$$$

→ Reply
»
4 years ago, # |
  Vote: I like it +122 Vote: I do not like it

The whole round is like an ingenuous play of MiFaFaOvO to get back his first place in the rating (though unsuccessful). Chinese people are really good at the art of war. My respect.

(please don't take this very seriously)

→ Reply
  • »
    »
    4 years ago, # ^ |
      Vote: I like it +22 Vote: I do not like it

    "There are two ways to go higher, self improvement, and sabotaging literally the entire community"

    He has made his choice.

    (Jk no offense pls :)

    → Reply
  • »
    »
    4 years ago, # ^ |
      Vote: I like it +11 Vote: I do not like it

    Seems his plan worked

    → Reply
»
4 years ago, # |
  Vote: I like it +41 Vote: I do not like it

Why the "non-empty" in C? I guess it has to do with printing the output but it reduced problem quality by like 30%.

→ Reply
  • »
    »
    4 years ago, # ^ |
      Vote: I like it -18 Vote: I do not like it

    could u plz xplain how to solve todays div2 C problem

    → Reply
    • »
      »
      »
      4 years ago, # ^ |
        Vote: I like it +35 Vote: I do not like it

      You do realize that I haven't read Div 2C?

      → Reply
      • »
        »
        »
        »
        4 years ago, # ^ |
          Vote: I like it -29 Vote: I do not like it

        ok..if someone other can help..bcz..i am very curious about it..that why my logic was wrong..even on sample test cases ..i know i was completely wrong.can not wait till editorial..xD

        → Reply
        • »
          »
          »
          »
          »
          4 years ago, # ^ |
            Vote: I like it +3 Vote: I do not like it

          rank 1 guys code is very clean to understand.

          → Reply
  • »
    »
    4 years ago, # ^ |
      Vote: I like it +39 Vote: I do not like it

    Might as well allow output of negative length to boost problem quality?

    → Reply
    • »
      »
      »
      4 years ago, # ^ |
        Vote: I like it +12 Vote: I do not like it

      Empty strings make sense, strings of negative length don't.

      → Reply
  • »
    »
    4 years ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    OH SHIT I missed it. Even my starting answer is an empty string. Luckily my code goes for large strings first when improving the cost in binsearch, so I got AC.

    → Reply
»
4 years ago, # |
  Vote: I like it 0 Vote: I do not like it

Is C checking if a cube like shape fits all of the given points by binary search / math?

→ Reply
  • »
    »
    4 years ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    Yes!But I am too weak at geometry,I come up with this idea quickly but don't know how to compute the area covered by several polygon...

    → Reply
    • »
      »
      »
      4 years ago, # ^ |
      ← Rev. 2   Vote: I like it +22 Vote: I do not like it

      Maintain the minimum and maximum values of $$$N$$$, $$$B$$$, and $$$N - B$$$.

      → Reply
      • »
        »
        »
        »
        4 years ago, # ^ |
          Vote: I like it 0 Vote: I do not like it

        Yes,but does it turn into a geometry problem??

        → Reply
        • »
          »
          »
          »
          »
          4 years ago, # ^ |
          ← Rev. 2   Vote: I like it 0 Vote: I do not like it

          Depends on your definition of "geometry" I guess. I didn't really use any "geometric" techniques. But yeah, you can visualize the things as hexagons.

          → Reply
        • »
          »
          »
          »
          »
          4 years ago, # ^ |
            Vote: I like it 0 Vote: I do not like it

          No need to compute the area. Just find a point inside it.

          → Reply
          • »
            »
            »
            »
            »
            »
            4 years ago, # ^ |
              Vote: I like it 0 Vote: I do not like it

            Yes,But how to compute the point inside it?

            → Reply
    • »
      »
      »
      4 years ago, # ^ |
        Vote: I like it +24 Vote: I do not like it

      Several polygons? I think it reduces to finding the minimum side length of the square in a shape like this that can fit all the points, which should be fairly easy. But I spent most of the contest thinking I had to minimize the sum of distances instead of the maximum, so I couldn't implement it in time.

      cube

      → Reply
  • »
    »
    4 years ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    Rather diamond-like (with top-left and bottom-right edges cut). Math seems awful there.

    → Reply
»
4 years ago, # |
← Rev. 2   Vote: I like it +4 Vote: I do not like it

How to solve Div2C? Also, where does the N^2 check all pairs approach go incorrect?

→ Reply
  • »
    »
    4 years ago, # ^ |
      Vote: I like it +6 Vote: I do not like it

    Final Answer can be at most $$$2 ^ {9}$$$, so just brute force for every possible value, and pick the smallest.

    → Reply
    • »
      »
      »
      4 years ago, # ^ |
        Vote: I like it 0 Vote: I do not like it

      Thanks.

      Do you think there a polynomial-time solution in case final answer is large?

      → Reply
      • »
        »
        »
        »
        4 years ago, # ^ |
          Vote: I like it +3 Vote: I do not like it

        I have O(n^2 log m), where m is the size of the largest input integer.

        Figure out how many leading zeroes we can get by picking greedily the pairs where the bitwise-and has the most leading zeroes, using O(n^2) time. The next bit has to be a 1 in any solution, so we can add that bit the the answer. Since the bit will be a 1 in any case, we can ignore is from now on: we set it to zero and find out again how many leading zeroes we can now get. Repeat until we can get all zeroes.

        https://codeforces.net/contest/1395/submission/89696109

        → Reply
    • »
      »
      »
      4 years ago, # ^ |
        Vote: I like it 0 Vote: I do not like it

      Your implementation is too elegant :)

      → Reply
  • »
    »
    4 years ago, # ^ |
    ← Rev. 3   Vote: I like it +12 Vote: I do not like it
    I solve this way
    → Reply
    • »
      »
      »
      4 years ago, # ^ |
        Vote: I like it 0 Vote: I do not like it

      Now I feel like an idiot,spend half an hour in implementing an dp solution..:/

      → Reply
      • »
        »
        »
        »
        4 years ago, # ^ |
          Vote: I like it 0 Vote: I do not like it

        Hey can we solve problem C it using Dp??
        I tried some approach but its giving wrong ans.

        → Reply
    • »
      »
      »
      4 years ago, # ^ |
        Vote: I like it 0 Vote: I do not like it

      Can you explain this approach?

      → Reply
      • »
        »
        »
        »
        4 years ago, # ^ |
          Vote: I like it 0 Vote: I do not like it
        My english is so bad.sorry.if you understand its my pleasure <3
        → Reply
    • »
      »
      »
      4 years ago, # ^ |
        Vote: I like it 0 Vote: I do not like it

      Very clean aproach, fix the answer and check there are N pairs a[i]&b[j] that has the same bit positions. I spend 1 hour until I reach my dp aproach...

      → Reply
»
4 years ago, # |
  Vote: I like it +24 Vote: I do not like it

Is 1C a geometry problem? I thought it just need to binary_search and compute the area covered by several polygon..

→ Reply
»
4 years ago, # |
  Vote: I like it 0 Vote: I do not like it

I passed the pretests in Div2B with a backtracking solution, how is that a thing?

→ Reply
»
4 years ago, # |
  Vote: I like it +13 Vote: I do not like it

Hints for $$$Div 2 E$$$ Please?

→ Reply
  • »
    »
    4 years ago, # ^ |
    ← Rev. 2   Vote: I like it +10 Vote: I do not like it

    Assume you have selected a tuple. Of course, you can now reduce the initial graph to one such that each node has out-degree of 1. Realise that for each node to be in a cycle in this resulting graph (and thus be possible to reach back to it in finite time), each node must also have an in-degree of 1.

    For each node (let it be Q), traverse over all nodes that point to it, let's call these set of nodes P. You can easily determine that for which value in the tuple will these nodes in P be pointing to Q. Also, this means that these values can not co exist in the tuple, since that would mean that multiple nodes point to Q in the resulting graph. Thus, you make a list of all these invalid combinations.

    Iterate over all K! tuples, and for each check if they consist of an invalid combination. If not, then it is valid.

    Sadly, was a bit late while implementing in contest :(

    Here's my AC code: https://codeforces.net/contest/1395/submission/89731507

    → Reply
    • »
      »
      »
      4 years ago, # ^ |
        Vote: I like it 0 Vote: I do not like it

      What's the upper bound of the number of those invalid combinations?

      → Reply
      • »
        »
        »
        »
        4 years ago, # ^ |
          Vote: I like it +16 Vote: I do not like it

        By an invalid combination, I mean pairs $$$((i, j), (k, l))$$$ such that value $$$j$$$ at index $$$i$$$ in the tuple, and value $$$k$$$ at index $$$l$$$ in the tuple, are incompatible. So the maximum number of such invalid combinations are $$$K!(K!+1) / 2$$$.

        You just have to store all these pairs in a $$$K*K*K*K$$$ matrix.

        → Reply
»
4 years ago, # |
  Vote: I like it +24 Vote: I do not like it

1250 not enough points for Div.2D

→ Reply
»
4 years ago, # |
← Rev. 3   Vote: I like it +8 Vote: I do not like it

Long Long in problem A just ruined my whole contest.I almost wasted 50min behind it.

→ Reply
»
4 years ago, # |
← Rev. 2   Vote: I like it +3 Vote: I do not like it

In Div2C, Sample test 3, what combinations get the final result as 147? The minimum I could get was 177.

→ Reply
  • »
    »
    4 years ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    I think you applied the dp approach. I did the same thing at first and waste a lot of time. just brute force the answer

    → Reply
  • »
    »
    4 years ago, # ^ |
    ← Rev. 3   Vote: I like it +1 Vote: I do not like it

    I think this was the combination

    Spoiler

    these are indices of the element from array a and b

    → Reply
  • »
    »
    4 years ago, # ^ |
    ← Rev. 2   Vote: I like it 0 Vote: I do not like it

    I also got the same error during the contest but now I understand. Well, let's say our solution consists of two parts, 1 and 2. I am not going to explain the second part assuming you already know it.

    But in the first part, u must be choosing a minimum c1 among all possible c1 and then doing part 2 of the solution but we are doing a mistake here. Actually, part 1 of the solution has n possibilities. Choosing minimum c1 is just one of the possibilities. A second possibility, for example, can be choosing a minimum c2 and then doing part 2 of the solution. Similarily choosing a minimum c3 among all possible c3 can be the third possibility and so on. So there are n possibilities.
    So the solution is O(n^3) and not O(n^2). wicardobeth

    → Reply
»
4 years ago, # |
  Vote: I like it +11 Vote: I do not like it

i will fst in d,One detail was not considered,shit.

→ Reply
»
4 years ago, # |
  Vote: I like it +8 Vote: I do not like it

That was implementationforces, I like that. Should be enough to become blue again :)

→ Reply
  • »
    »
    4 years ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    What was the solution to D?

    → Reply
    • »
      »
      »
      4 years ago, # ^ |
      ← Rev. 2   Vote: I like it +10 Vote: I do not like it

      Greedy.

      Edit, more understandable: We devide $$$a[]$$$ into set $$$a[]$$$ and $$$b[]$$$, where all $$$b[i]<=m$$$ and all $$$a[i]>m$$$

      Sort both sets biggest element first.

      Then choose consecutevly to use i elements from a[]. Foreach i we can calculate how much elements from b[] can be used then, and using prefix-sum calc in O(1) the funfactor of that i.

      → Reply
      • »
        »
        »
        »
        4 years ago, # ^ |
          Vote: I like it +4 Vote: I do not like it

        If only I had two more minutes to fix my implementation

        → Reply
      • »
        »
        »
        »
        4 years ago, # ^ |
          Vote: I like it 0 Vote: I do not like it

        Figured that, but couldn't implement. Thanks, will wait till I can see your submission.

        → Reply
  • »
    »
    4 years ago, # ^ |
      Vote: I like it +3 Vote: I do not like it

    I am used to see you in blue ;)

    → Reply
»
4 years ago, # |
  Vote: I like it +1 Vote: I do not like it

How to solve div2 D?

→ Reply
  • »
    »
    4 years ago, # ^ |
      Vote: I like it +6 Vote: I do not like it

    1) Make 2 lists: one of all greater than m (let us call it gr), second all less than m(let us call it sm). 2) Sort gr reverse and sm normally. 3) Now compare highest unused of gr with sum of d consecutive of sm. If gr highest is greater than simply add that to ans(in other words muzzle d smallest elements of sm). Take care of these things: 1) Do the above only till you have atleast 1 element left in gr, cause you can add it in the end. 2) If you encounter condition like less than d elements remaining in sm, then dont compare with gr greatest element as you can add all remaining along with greatest, if greatest goes in the end. 3) In the end try to put all remaining elements of gr in the end of the permutation seperated by d elements each. For details see the code: 89725326 If I fail system test ignore this comment :D

    → Reply
»
4 years ago, # |
← Rev. 2   Vote: I like it 0 Vote: I do not like it

Can anyone please tell me what's wrong in my code? Problem A https://ide.codingblocks.com/s/308173

→ Reply
»
4 years ago, # |
  Vote: I like it +7 Vote: I do not like it

Can someone tell me why this solution to 1C is wrong?89722745

I can't find the mistake :(

→ Reply
  • »
    »
    4 years ago, # ^ |
      Vote: I like it +7 Vote: I do not like it

    Oh now I find it

    I forgot one line if(zmin>zmax) return 0; :(

    → Reply
»
4 years ago, # |
  Vote: I like it +1 Vote: I do not like it

Boboniu stands nowhere in front of Binod

→ Reply
»
4 years ago, # |
  Vote: I like it +3 Vote: I do not like it

Any Idea of Test 16, Div1 B ?

→ Reply
»
4 years ago, # |
  Vote: I like it +8 Vote: I do not like it

How to solve Div-1 B??

→ Reply
»
4 years ago, # |
  Vote: I like it 0 Vote: I do not like it

Need help, why did my solution for Div2B did not pass ? I have used dfs- https://codeforces.net/contest/1395/submission/89714645

→ Reply
»
4 years ago, # |
← Rev. 2   Vote: I like it 0 Vote: I do not like it

Problem A: What is the problem with my this python code?

I decremented min(r, g, b) from r , g, b. Then count how many of them are odd from r, g, b, w. If number of odds > 1 the answer is "No", else "Yes".

Sorry for bad English.

→ Reply
  • »
    »
    4 years ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    What if input is 2 5 7 3. Now if u decrement r g b by 2, then r b g will be 0 3 5 and w will be 9.

    No.of odds will be 3 (0 3 5 9)..so it return "No"

    But if I remove only 1 from r g b then r g b will be 1 4 6 and w will be 6. (1 4 6 6), we can form a palindrome.

    Decrementing min(r,g,b) is wrong step.

    → Reply
»
4 years ago, # |
← Rev. 2   Vote: I like it +3 Vote: I do not like it

Can someone give a small hint for 'B'?

As far as I understand:
1) The requirement "should be able to return back" is equivalent to:
= edges that satisfy Ci-rules should form loops
= every vertex should be a part of a loop
= every vertex should have an active edge coming in — this is what we need to check
2) k <= 9 so 1*2*3*4*5*6*7*8*9 = 362880 total sequences of Ci.
If we were able to somehow check whether a specific sequence is good fast, than we could just enumerate all sequences and count the good ones.

Am I correct with (1) and (2)? If no, a small hint would be appreciated.

→ Reply
  • »
    »
    4 years ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    Isn't the total number of sequences = k^9?

    → Reply
    • »
      »
      »
      4 years ago, # ^ |
        Vote: I like it 0 Vote: I do not like it

      No, since c_i <= i, not c_i <= k

      → Reply
    • »
      »
      »
      4 years ago, # ^ |
        Vote: I like it +8 Vote: I do not like it

      No, because 1 <= C[i] <= i

      Remember, C[i] — is the index of the edge you would take if there are 'i' outcoming edges. This index can't be say 7 if there are only 3 outbound edges.

      → Reply
  • »
    »
    4 years ago, # ^ |
    ← Rev. 2   Vote: I like it +13 Vote: I do not like it

    Sufficient and necessary condition is that each vertex has exactly one incoming edge. Everything else in your (1) follows.

    I build a bruteforce on top of that with

    P.S. It's pretty standard Markdown, so use two spaces before newline
    like this

    → Reply
    • »
      »
      »
      4 years ago, # ^ |
        Vote: I like it 0 Vote: I do not like it

      How do you prove that?

      → Reply
      • »
        »
        »
        »
        4 years ago, # ^ |
          Vote: I like it 0 Vote: I do not like it

        Necessary is obvious — no incoming edge, no cycle for that vertex.

        Sufficient: start from vertex and start moving. At some point you will have to return to starting vertex.

        → Reply
      • »
        »
        »
        »
        4 years ago, # ^ |
          Vote: I like it +5 Vote: I do not like it

        If a node $$$x$$$ has no incoming edge, it's impossible to reach $$$x$$$ starting from $$$x$$$.
        If a node $$$x$$$ has indegree $$$\geq 2$$$, let $$$a, b$$$ be two nodes such that edges $$$(a, x)$$$ and $$$(b, x)$$$ exist. But if you start from $$$x$$$ you can't reach both $$$a$$$ and $$$b$$$ (wlog, if you reach $$$a$$$, you return to $$$x$$$ and you haven't visited $$$b$$$); in this case, if you start from $$$b$$$ you reach $$$x$$$ at the next step and you can't go back to $$$b$$$.

        → Reply
        • »
          »
          »
          »
          »
          4 years ago, # ^ |
            Vote: I like it 0 Vote: I do not like it

          what if there is an edge (a, b) along with (a, x) and (b, x) ? Then can't it be the case such that: x -> ......... -> a -> b -> x. ?

          → Reply
          • »
            »
            »
            »
            »
            »
            4 years ago, # ^ |
              Vote: I like it +8 Vote: I do not like it

            There can't be both (a,b) and (a,x)

            → Reply
    • »
      »
      »
      4 years ago, # ^ |
        Vote: I like it 0 Vote: I do not like it

      Then how to check if some node has no incoming edge?

      → Reply
      • »
        »
        »
        »
        4 years ago, # ^ |
          Vote: I like it +11 Vote: I do not like it

        Each assignment $$$c_i = j$$$ will guarantee incoming edges for some number of vertices. If you sum this up over all assignments you make, just check that it adds up to $$$n$$$.

        The issue is that this can double count, however note that each vertex should have exactly one incoming edge, not more, so each assignment $$$c_i = j$$$ will exclude some other list of assignments $$${c_{i'} = j'}$$$. Just make sure to never make two conflicting assignments like this.

        → Reply
      • »
        »
        »
        »
        4 years ago, # ^ |
        ← Rev. 2   Vote: I like it +3 Vote: I do not like it

        You have n edges, so it's enough to check that no vertex has two incoming edges. Precompute 'bad' pairs (i, c_i), (j, c_j), such that some vertex will have two incoming edges.

        Simple bruteforce after that.

        → Reply
  • »
    »
    4 years ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    You are correct with both (1) and (2).

    Here's a good hint. Imagine that for a certain tuple of C, we get the edges 1-2,2-1,3-4,4-3. This is equivalent to saying that nodes [1,2,3,4]->[2,1,3,4] which is a permutation! How can you use this insight to solve the problem in general?

    → Reply
    • »
      »
      »
      4 years ago, # ^ |
        Vote: I like it +8 Vote: I do not like it

      Agnimandur kilotaras thanks! Solved!

      The core idea is to:
      → Reply
  • »
    »
    4 years ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    I thought the exact same thing and could not determine how to check every possible sequence in an optimized way.

    → Reply
»
4 years ago, # |
  Vote: I like it 0 Vote: I do not like it

Does anyone know if there is a testcase on Div1B which breaks the hashing solution for the MOD 1e9+7?

→ Reply
»
4 years ago, # |
  Vote: I like it +3 Vote: I do not like it

Very nice problemset, reminded me to practice.

→ Reply
»
4 years ago, # |
← Rev. 2   Vote: I like it +99 Vote: I do not like it

[deleting this because it came out wrong. I'm referring to Chinas efforts in testing all of Wuhan, which is amazing, not them experimenting and creating the virus (which I think is false).

I can't delete this, so if some CF mod could for me that'd be great.

→ Reply
»
4 years ago, # |
  Vote: I like it +103 Vote: I do not like it

the use of "muzzle" in problem A felt rather odd.

→ Reply
»
4 years ago, # |
  Vote: I like it +1 Vote: I do not like it

tourist RIP rating

→ Reply
»
4 years ago, # |
  Vote: I like it +4 Vote: I do not like it

FAILED TEST 31 IN C? GOING BACK TO DIV4.

→ Reply
»
4 years ago, # |
← Rev. 4   Vote: I like it +74 Vote: I do not like it

I am surprised by the number of high-rated coders who do not realize that in Div1B, the in-degree of each node can be up to O(n), and iterating through pairs of nodes which go into a particular node can take quadratic time in the worst case.

I am also surprised that there was no pretest against such solutions. But I think this is fair. Contestants are responsible for ensuring that their solutions' time complexities are acceptable.

→ Reply
»
4 years ago, # |
  Vote: I like it +41 Vote: I do not like it

Pretests were way too weak, just look at the amount of failed submissions in Div1 ...

→ Reply
  • »
    »
    4 years ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    My finish in Div1 went from 556 to 366 because of failed submissions :D

    → Reply
»
4 years ago, # |
  Vote: I like it +30 Vote: I do not like it

What the hell were you guys thinking. I don't usually get mad, but now I am mad. This contest honestly has made me not want to do cf anymore

→ Reply
»
4 years ago, # |
← Rev. 2   Vote: I like it +22 Vote: I do not like it

Absolutely disappointed

→ Reply
»
4 years ago, # |
  Vote: I like it +37 Vote: I do not like it

my disappointment is immeasurable and my day is ruined

→ Reply
»
4 years ago, # |
  Vote: I like it 0 Vote: I do not like it

My only question is why?

→ Reply
»
4 years ago, # |
  Vote: I like it +4 Vote: I do not like it

Weak Pretests :(

→ Reply
»
4 years ago, # |
  Vote: I like it +35 Vote: I do not like it

WHYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

→ Reply
  • »
    »
    4 years ago, # ^ |
      Vote: I like it +57 Vote: I do not like it

    You could've said all of this in 1 comment?

    → Reply
  • »
    »
    4 years ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    AYAYAYAYAAYAYAYAYAAAYAYAYAYA AYA AYAY A YA YA YA PSY PSY PSY PSY DUCKDUCKDUCKDUCKKK

    → Reply
»
4 years ago, # |
  Vote: I like it +29 Vote: I do not like it

So many main tests failing

→ Reply
»
4 years ago, # |
  Vote: I like it +42 Vote: I do not like it

Worse pretests ever.

→ Reply
»
4 years ago, # |
← Rev. 2   Vote: I like it -9 Vote: I do not like it

89726386 here is my solution for D, it for some reason gave no output nor any errors. its probably a silly mistake, but i cant find it. could anyone help please?

→ Reply
»
4 years ago, # |
  Vote: I like it +13 Vote: I do not like it

I was so happy after giving the round but Wrong answer on test 31.

→ Reply
»
4 years ago, # |
  Vote: I like it 0 Vote: I do not like it

Going to Specialist~~~

→ Reply
»
4 years ago, # |
  Vote: I like it +3 Vote: I do not like it

Even though the round was great with short (and to the point although a bit unclear initially) problem statements, I found the pretests to be weak..lots of solutions are failing the system tests.

→ Reply
»
4 years ago, # |
  Vote: I like it -15 Vote: I do not like it

Very weak pretests for Div2 D , even the most simple side cases werent inside the tests.

→ Reply
  • »
    »
    4 years ago, # ^ |
      Vote: I like it -11 Vote: I do not like it

    literally missed one case: if the elements are all > m..... got it accepted instantly after

    → Reply
»
4 years ago, # |
  Vote: I like it +2 Vote: I do not like it

Am I the only one who solved Div2 C with dp? 89700923

→ Reply
»
4 years ago, # |
← Rev. 5   Vote: I like it +303 Vote: I do not like it

→ Reply
»
4 years ago, # |
  Vote: I like it +70 Vote: I do not like it
Mood
→ Reply
»
4 years ago, # |
  Vote: I like it +5 Vote: I do not like it

( ཀ ʖ̯ ཀ):

→ Reply
»
4 years ago, # |
  Vote: I like it 0 Vote: I do not like it

Alot of solutions are getting rejected after passing pretests i think

→ Reply
»
4 years ago, # |
  Vote: I like it 0 Vote: I do not like it

Why doing nested ternary search is correct in problem C?

→ Reply
»
4 years ago, # |
  Vote: I like it 0 Vote: I do not like it

Why is brute force O(n*m) failing for div2 C ?

→ Reply
  • »
    »
    4 years ago, # ^ |
      Vote: I like it +1 Vote: I do not like it

    No it doesn't. My O(2^9 * n * m) worked fine.

    I can't see a div2 C submission on your profile. Can you share your code

    → Reply
    • »
      »
      »
      4 years ago, # ^ |
      ← Rev. 2   Vote: I like it +1 Vote: I do not like it 
      #include <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define mod 1000000007

int main(){
    int t=1;
    while(t--){
        ll n,m;
        cin>>n>>m;
        ll a[n];
        for(int i=0;i<n;++i)cin>>a[i];
        ll b[m];
        for(int i=0;i<m;++i)cin>>b[i];
        vector<ll> c;
        for(int i=0;i<n;++i){
            ll minn = INT_MAX;
            for(int j=0;j<m;++j){
                minn = min(minn,(a[i]&b[j]));
            }
            c.push_back(minn);
        }
        ll ans = 0;
        for(int i=0;i<c.size();++i){
            ans = (ans|c[i]);
        }
        cout<<ans;
    }
    return 0;
}
      → Reply
      • »
        »
        »
        »
        4 years ago, # ^ |
          Vote: I like it 0 Vote: I do not like it

        It's not always optimal to take $$$min(a_i$$$ & $$$b_j)$$$. Which bits you are taking matters.

        → Reply
        • »
          »
          »
          »
          »
          4 years ago, # ^ |
            Vote: I like it 0 Vote: I do not like it

          can you elaborate with an example please?

          → Reply
          • »
            »
            »
            »
            »
            »
            4 years ago, # ^ |
              Vote: I like it +1 Vote: I do not like it

            Let's say the answer has a particular bit set to 1, so it means that atleast one of c1,c2...cn will definitely have that particular bit set to 1. In that case we don't need to try and set that bit to 0 for other values in c1,c2...cn by doing the AND operation since the answer would still have that bit set to 1. Guess that makes it clear why taking the minimum won't help here, it's the bits you are taking that matter.

            → Reply
            • »
              »
              »
              »
              »
              »
              »
              4 years ago, # ^ |
                Vote: I like it +1 Vote: I do not like it

              Got it, thanks for your efforts :)

              → Reply
          • »
            »
            »
            »
            »
            »
            4 years ago, # ^ |
              Vote: I like it 0 Vote: I do not like it 
            3 2
13 5 8
4 9

            Your code's output is $$$5$$$. But the answer is (13 & 4) | (5 & 4) | (8 & 4) = 4.

            → Reply
»
4 years ago, # |
  Vote: I like it +5 Vote: I do not like it

When you're feeling good after a contest getting A B and C, and it's looking like you're going to gain a lot of rating, and it just all gets crushed by failed system testing due to weak pretests :(

→ Reply
»
4 years ago, # |
  Vote: I like it -16 Vote: I do not like it

PLEASE MAKE PRETESTS STRONG !

→ Reply
»
4 years ago, # |
← Rev. 3   Vote: I like it 0 Vote: I do not like it

My div2A failed test 10 ,and div2B failed test 39 :( Why these tests were not used in contest ?

→ Reply
»
4 years ago, # |
  Vote: I like it +6 Vote: I do not like it

Everyone to their before system testing score: Those son of bitch lied to me!

→ Reply
»
4 years ago, # |
  Vote: I like it +3 Vote: I do not like it

in B div1 I don't get why the 9!*9*9 ~= 3e7 should not pass the tests but with some optimizations it should pass. the 1s time limit is so stupidly chosen.

→ Reply
  • »
    »
    4 years ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    Your code seems to have complexity not 9!*9*9, but way more. If there is a vertex, with $$$O(n)$$$ incoming edges it will be quadratic.

    → Reply
    • »
      »
      »
      4 years ago, # ^ |
        Vote: I like it +10 Vote: I do not like it

      Ow right. but it need just a sort and unique and it will be ok. pretest must include these kind of tests i think.

      → Reply
»
4 years ago, # |
  Vote: I like it -20 Vote: I do not like it

Pretests just ruined this (I don't think it is supposed to be that many "failed on systests"). Your contest suck!

→ Reply
»
4 years ago, # |
  Vote: I like it +14 Vote: I do not like it

Over 850+ System test failed in problem C. Including me :(

→ Reply
»
4 years ago, # |
  Vote: I like it 0 Vote: I do not like it

The following submission https://codeforces.net/contest/1394/submission/89716136 gives runtime error on pretest 3 on the judge server but it works fine on my local pc and on ideone when tested now. Can anyone tell me why this is going wrong? Is it because codeforces uses the CLANG compiler?

→ Reply
  • »
    »
    4 years ago, # ^ |
    ← Rev. 2   Vote: I like it +8 Vote: I do not like it

    It will give runtime error when less.size() == 1, because you are accessing less.size() - 2 (-1) but unsigned int can't be negative so it becomes UINT_MAX.

    → Reply
»
4 years ago, # |
  Vote: I like it +10 Vote: I do not like it

The pretests were kinda weak:((

→ Reply
»
4 years ago, # |
  Vote: I like it -15 Vote: I do not like it

I think it's not just that getting faild on sample tests decrease our point, because sample tests are for testing our code. In this contest, I get 2 or 3 WA's on pretest 2 of problem A that is a sample test, and it decreased huge points from my solution.

→ Reply
»
4 years ago, # |
  Vote: I like it +1 Vote: I do not like it

I did Div2 C like this: https://codeforces.net/contest/1395/submission/89706416 with O(n^2) complexity but I'm not able to think of a proof exactly for why this is right, can someone help me out?

→ Reply
  • »
    »
    4 years ago, # ^ |
      Vote: I like it +3 Vote: I do not like it

    I too had solved this problem in the exact same way and I'm currently unable to guess why this is working. https://codeforces.net/contest/1395/submission/89730303

    In fact I also tried another small tweak in this algo but this submission passed all the tests too! https://codeforces.net/contest/1395/submission/89734201

    I wonder if the problem lies in the testcases

    → Reply
  • »
    »
    4 years ago, # ^ |
    ← Rev. 2   Vote: I like it +3 Vote: I do not like it

    In the first loop, $$$mn$$$ is the minimum $$$a_i$$$ & $$$b_j$$$ for each $$$i$$$ and $$$mx$$$ is the maximum $$$mn$$$. This indicates that after performing $$$OR$$$, the answer will be at least $$$mx$$$ and it can be more than $$$mx$$$ if other $$$mn$$$ has some on bits which are off in $$$mx$$$. In the second loop, we are just turning on those off bits of $$$mx$$$ to find the final result.
    UPD: This solution doesn't provide correct output for all cases.

    → Reply
»
4 years ago, # |
← Rev. 3   Vote: I like it +26 Vote: I do not like it

Weak pretests! and testers asking for upvotes

me

→ Reply
»
4 years ago, # |
  Vote: I like it -13 Vote: I do not like it

→ Reply
»
4 years ago, # |
  Vote: I like it +163 Vote: I do not like it

To not keep you waiting, the ratings updated preliminarily. In a few hours, I will remove cheaters and update the ratings again!

→ Reply
»
4 years ago, # |
  Vote: I like it +3 Vote: I do not like it

My brute force solution passed Div1B.

Can anyone hack it? :)

this

→ Reply
»
4 years ago, # |
  Vote: I like it 0 Vote: I do not like it

Hi everyone, I was stuck in Div2-C for more than half an hour. People say if you are stuck for more than half an hour, you shouldn't try more and just wait for the editorial. So, I gave up. Later when only 15 mins were left, the idea struck my mind. And I couldn't implement within those 15 mins. But after the contest, I submitted and it was correct. So, how should I decide whether I should give up or not?

→ Reply
  • »
    »
    4 years ago, # ^ |
      Vote: I like it +26 Vote: I do not like it

    Never give up in a live contest. Try the other problems, and keep thinking until there's no time left.

    → Reply
»
4 years ago, # |
  Vote: I like it 0 Vote: I do not like it

It's so unfair when pretests are too weak, especially in problem C, D. Luckily, I've passed the system test, still, I would say it's not fair.

→ Reply
»
4 years ago, # |
  Vote: I like it +7 Vote: I do not like it

code
Why this solution of O(n*m) for problem C is giving TLE on test 10

→ Reply
»
4 years ago, # |
  Vote: I like it 0 Vote: I do not like it

Weak pretests in D. It hurts.

→ Reply
»
4 years ago, # |
  Vote: I like it 0 Vote: I do not like it

Oh... I get it. They didn't write "we tried to make strong pretests" so they aren't.

→ Reply
»
4 years ago, # |
  Vote: I like it 0 Vote: I do not like it

Can anyone tell why my code for div2 C is giving TLE. code

→ Reply
»
4 years ago, # |
  Vote: I like it +1 Vote: I do not like it

During the contest in Div.2 C, the verdict was pretests passed and after system testing, it became wrong answer on test case 9. Can someone explain me why?

→ Reply
  • »
    »
    4 years ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    During the contest your submission is only tested on few tests. The rest of the tests are processed after the contest.

    → Reply
»
4 years ago, # |
  Vote: I like it +1 Vote: I do not like it

Failed system test in problem C. And finding out that it can be solved using brute force. Just disastrous.

→ Reply
»
4 years ago, # |
  Vote: I like it +9 Vote: I do not like it

http://codeforces.net/contest/1394/submission/89728986 A hack for Div1 B

I enumerated all incoming-edge pairs of every vertex in this solution, which has a complexity of O(n^2) in the worst case.

However, it got Accepted when I submitted after contest :(

By the way, I improved my solution and resubmit for 3 times during the contest in order to fix the bug. And, my new code Failed System Test because of another mistake TAT

Wish for stronger system tests :(

→ Reply
  • »
    »
    4 years ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    Contest was amazing but Testers ruined the contest

    → Reply
»
4 years ago, # |
  Vote: I like it +5 Vote: I do not like it

Ok, I have been trying to debug my code for Div1B for more than 2 hrs, but I am still not able to find why this is failing. Its just a simple brute force. Can anyone point out where I am going wrong?

→ Reply
  • »
    »
    4 years ago, # ^ |
      Vote: I like it +5 Vote: I do not like it

    Your solution undercounted because "cnt == n — 1" ... a valid solution can result in more than one cycles being formed.

    Here is a simple test case to illustrate:

    4 8 9 1 2 1 2 1 3 3 4 2 4 3 4 1 3 10 3 1 30 2 4 20 4 2 40

    The answer is 362880 (equals to 9! which means any valid combination works). Your code ouputs zero.

    → Reply
    • »
      »
      »
      4 years ago, # ^ |
        Vote: I like it 0 Vote: I do not like it

      Ok, I got it. I assumed it will always be one big cycle. My bad, I was mislead by the test cases. Thanks a lot.

      → Reply
»
4 years ago, # |
  Vote: I like it 0 Vote: I do not like it

https://codeforces.net/contest/1395/submission/89733714

Can anyone help me out with Task A div 2

→ Reply
  • »
    »
    4 years ago, # ^ |
    ← Rev. 2   Vote: I like it 0 Vote: I do not like it

    Removing the min from r,b,g is not always correct. If you have more than 1 odd number you only need to remove 1 from r,b,g and add it to w. And if you still have more than 1 odd number the answer is no.

    → Reply
»
4 years ago, # |
  Vote: I like it 0 Vote: I do not like it

What is test case 16 of div2 D can anyone help:(

→ Reply
»
4 years ago, # |
  Vote: I like it 0 Vote: I do not like it

System tests ate my rating!

→ Reply
»
4 years ago, # |
  Vote: I like it -8 Vote: I do not like it

int main() { ios_base :: sync_with_stdio(false); cin.tie(NULL); auto startTime = curTime();

int n,m;
cin>>n>>m;

vi a(n);
vi b(m);
rep(i,n) cin>>a[i];
rep(i,m) cin>>b[i];

ll mx=0;
rep(i,n)
{
    ll mn = inf;
    rep(j,m)
    {
        mn = min(mn, a[i]&b[j]);
    }
    mx = max(mx,mn);
}
ll ans=0;
rep(i,n)
{
    ll mn = inf;
    rep(j,m)
    {
        mn = min(mn, mx|(a[i]&b[j]));
    }
    mx=mn;
    ans = max(ans,mn);
}

cout<<ans<<endl;

Can anyone explain the logic of this solution? I saw same thing in many accepted solutions.

→ Reply
  • »
    »
    4 years ago, # ^ |
    ← Rev. 2   Vote: I like it +3 Vote: I do not like it

    In the first loop, $$$mn$$$ is the minimum $$$a_i$$$ & $$$b_j$$$ for each $$$i$$$ and $$$mx$$$ is the maximum $$$mn$$$. This indicates that after performing $$$OR$$$, the answer will be at least $$$mx$$$ and it can be more than $$$mx$$$ if other $$$mn$$$ has some on bits which are off in $$$mx$$$. In the second loop, we are just turning on those off bits of $$$mx$$$ to find the final result.
    UPD: This solution doesn't provide correct output for all cases.

    → Reply
    • »
      »
      »
      4 years ago, # ^ |
        Vote: I like it 0 Vote: I do not like it

      How can you say that answer will be atleast mx?

      → Reply
      • »
        »
        »
        »
        4 years ago, # ^ |
          Vote: I like it 0 Vote: I do not like it

        Because $$$OR$$$ of $$$x$$$ and $$$y$$$ is never less than $$$x$$$ or $$$y$$$. That's why after performing $$$OR$$$, the answer will be at least $$$mx$$$.

        → Reply
      • »
        »
        »
        »
        4 years ago, # ^ |
          Vote: I like it 0 Vote: I do not like it

        Submissions with this solution is getting uphacked.

        → Reply
    • »
      »
      »
      4 years ago, # ^ |
      ← Rev. 2   Vote: I like it 0 Vote: I do not like it

      This solution seems to have a wrong output on Test 32

      → Reply
      • »
        »
        »
        »
        4 years ago, # ^ |
          Vote: I like it 0 Vote: I do not like it

        Which test case?

        → Reply
        • »
          »
          »
          »
          »
          4 years ago, # ^ |
          ← Rev. 3   Vote: I like it 0 Vote: I do not like it

          On Test 32, which inputs : 54 12 435 115 422 469 19 176 292 395 273 217 160 343 277 500 124 360 315 59 2 477 465 138 175 392 146 438 4 96 231 182 257 183 51 117 430 125 167 107 238 66 236 96 319 62 365 186 83 414 166 198 224 129 60 195 327 62 310 278 300 228 295 113 502 320 463 223

          we should output 63, but this code runs with an answer 62 .

          → Reply
»
4 years ago, # |
  Vote: I like it +7 Vote: I do not like it

Problems from this round are really nice ones! Maybe the pretests could have been stronger. My friend's code for problem Div1C is $$$O(n^2*log(n))$$$ and it passed pretests. I'm not sure if it is a good idea to use such weak pretests for CF rounds. Besides that, this was an interesting round. I enjoyed thinking Div1B. Thank you for your effort! :)

→ Reply
»
4 years ago, # |
← Rev. 2   Vote: I like it +132 Vote: I do not like it

In 1394C - Boboniu and String, there was no test with a string longer than $$$200\,000$$$ (while the limit was $$$500\,000$$$). People could use too small arrays or iterate/binarySearch in much smaller range. Come on, doesn't Polygon warn you automatically if you don't hit some min/max value?

My sad story is that I noticed my own mistake during a contest, resubmitted and thus dropped 5th -> 20th place. My first code gets AC though, and only now I uphacked it with one long string.

→ Reply
»
4 years ago, # |
  Vote: I like it +3 Vote: I do not like it

I develop app , CP helps me build logic and thinking capability.

→ Reply
»
4 years ago, # |
← Rev. 2   Vote: I like it -18 Vote: I do not like it

Thanks for the contest! But a little overkill B :( Many people complain about pretests, but I think pretest were actually good (upd: good means not strong, good — finally allows to hack).

→ Reply
  • »
    »
    4 years ago, # ^ |
      Vote: I like it -8 Vote: I do not like it

    Dude, are you nuts ? How can you call the pretests strong ? Every other solution is failing in system testing. Many solutions even failed A and B in Div 2

    → Reply
    • »
      »
      »
      4 years ago, # ^ |
        Vote: I like it 0 Vote: I do not like it

      I know it is hard skill to read, but try it again. I said "good", I did not say "strong". Also "pretests" is not the same as "tests". And good, because hacking this round was not useless feature again. But I was talking about Div 1. Div 2 A or B probably should have all tests in pretests.

      → Reply
»
4 years ago, # |
  Vote: I like it -34 Vote: I do not like it

Is the guy Binod?

→ Reply
»
4 years ago, # |
← Rev. 2   Vote: I like it +10 Vote: I do not like it

Any reason why the multiple testcases within a single test format was excluded today? Believe recent rounds had fairly less systest fails cause of that. Might be difficult / annoying for both parties but is still the better & safer choice imo

→ Reply
  • »
    »
    4 years ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    Yes and also when I am going through the test cases it seems that the pretests were just randomly made safeguarding the actual main tests.

    → Reply
    • »
      »
      »
      4 years ago, # ^ |
        Vote: I like it 0 Vote: I do not like it

      How do you know which of those tests were pretests ?

      → Reply
      • »
        »
        »
        »
        4 years ago, # ^ |
          Vote: I like it 0 Vote: I do not like it

        Just before system tests try using side bar filters present try looking for maximum test on which most of the submissions failed, that number is a good idea of the boundary condition for number of pretest. You have to do these stuffs during the contest.

        → Reply
        • »
          »
          »
          »
          »
          4 years ago, # ^ |
            Vote: I like it 0 Vote: I do not like it

          Ohh..so u mean out of the N tests that are visible now (after system testing), the pretests are the first K tests ?

          I initially thought that pretests are randomly shuffled in the system tests.

          → Reply
  • »
    »
    4 years ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    Your profile picture paired with your comment is perfected :o

    → Reply
»
4 years ago, # |
  Vote: I like it +3 Vote: I do not like it

Auto comment: topic has been updated by sshwyR (previous revision, new revision, compare).

→ Reply
»
4 years ago, # |
  Vote: I like it +8 Vote: I do not like it

Thank You for the Contest, I have finally become candidate master.

→ Reply
»
4 years ago, # |
  Vote: I like it 0 Vote: I do not like it

In div2 A problems, I found my code is wrong.

input > 

1
1 2 2 4

In the above test case, The answer is 'Yes' but my code prints 'No'. 89667141

(If the test case is added, Will the standings be affected?)

→ Reply
»
4 years ago, # |
  Vote: I like it 0 Vote: I do not like it

When will the editorials come out?

→ Reply
»
4 years ago, # |
  Vote: I like it 0 Vote: I do not like it

It is not easy to prepare a round in CF. This round is good except the test cases(not pretest but all test cases in some problems). For instance,in Div.2 C,many participant failed on test 31 which was emerged during the hacking phase. Except that,I am still greatful for the effort of writers and coordinators.

→ Reply
»
4 years ago, # |
  Vote: I like it 0 Vote: I do not like it

https://codeforces.net/contest/1395/submission/89766628 This is my answer to div2 E question. But there is a problem. This program cannot be guaranteed to be accepted. So here comes the question. If the program is submitted multiple times during the competition and all pass the pre-test. Which program will you use to run the system test? It is the last submitted program. Or do all programs that pass the pre-test run a system test?

→ Reply
  • »
    »
    4 years ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    The last correct submission is judged always.

    → Reply
    • »
      »
      »
      4 years ago, # ^ |
        Vote: I like it 0 Vote: I do not like it

      OK, thanks a lot. Therefore, if a random algorithm is to be guaranteed to be accepted, it has to run multiple times. It’s just that there is a theoretically wrong probability, even if it’s small

      → Reply
»
4 years ago, # |
  Vote: I like it 0 Vote: I do not like it

I'm unable to understand the editorial for div2- C. Can someone explain to me clearly?

→ Reply
  • »
    »
    4 years ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    If X|Y = X then no new set bits are introduced in Y. If we have to check whether the answer is atleast X then for every i there should be a j such that (a[i]&b[j])|x = x, we iterate over x and find out which is minimum

    → Reply
»
4 years ago, # |
  Vote: I like it 0 Vote: I do not like it

editorial?

→ Reply
»
4 years ago, # |
  Vote: I like it 0 Vote: I do not like it

How to prove that the function is unimodal in Div1C(Div2F)?

Thanks

→ Reply
»
4 years ago, # |
  Vote: I like it 0 Vote: I do not like it

Btw did nobody notice that div1D had pretests = systests? There were some "WA on pretest 71" during the contest.

→ Reply
»
4 years ago, # |
  Vote: I like it 0 Vote: I do not like it

In problem 2 the vertical coordinate was X and the horizontal one was Y. Why?

→ Reply
»
4 years ago, # |
  Vote: I like it 0 Vote: I do not like it

In Div2 problem E, why aren't the tuple's length equal with the cycle's edge number?

Like in example one, one of the cycle is from node 1 to 2, to 4, to 3, then back to 1 with 4 edges, so the tuple should be (1,1,3,1), right? But the answer tuples are (1,1,3) and (1,2,3). Is there something that I missed about this problem's statement?

→ Reply
»
4 years ago, # |
← Rev. 2   Vote: I like it 0 Vote: I do not like it

code why this code for problem C won't get TLE

→ Reply