Анонс Looksery Cup 2015

9 лет назад, # ^ |

← Rev. 3 →

Please ignore this comment. It initially contained an incorrect argument. I thank kingofnumbers for correcting me.

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9 лет назад, # ^ |

However, here if someone takes a top 200 place(or any other place for that matter) then had they participated with their official account they would have taken the same place so it should't really have any affect on anyone else.

Well, you forgot that if they taken part with fake account then the rating of that account will increase more than if they taken part with their real accounts and this will make others get less higher ratings

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9 лет назад, # ^ |

Hmm. I was under the impression that ones rating change was only a function of ones position and ones initial rating but it turns out that was incorrect.

Sorry about that.

It would still be nice if people were allowed to have a choice in the text that got printed though.

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9 лет назад, # ^ |

Nooo. Imagine two Div1 contests. One where 500 violets take part and one where 500 reds take part. Your possibility wouldn't make sense :P.

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StoneCold_

9 лет назад, # ^ |

Even though this comment is not relevant to the discussion but I would like to inform you that The rating calculation for unrated people is a little different . Unrated contestants have an initial expected rank of n / 2 where n is the number of people taking part in the contest. So in a contest consisting of 499 red coders and 1 unrated coder if the unrated coder gets a rank worse than 250 then his rating would drop.

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pk.

9 лет назад, # ^ |

I guess here he was not referring to rating or anything.

He just mentioned "to get T-shirts name of their choice"

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I_Love_Tina

9 лет назад, # |

+171

nice t-shirts !!

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shakil_AUST

9 лет назад, # |

-90

Can you please give me the t-shirt for nothing :(

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inakolithsihsa

9 лет назад, # ^ |

+62

stop begging...start winning :P

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shakil_AUST

9 лет назад, # ^ |

← Rev. 2 →

-36

I am just trying to comment something funny but alas it becomes too funny to give up vote :D

Thanx to Looksery and Codeforces team for that opportunity and prizes are very much tempting . Hope to have a great round with high rating everyone :) May the best win :)

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suxrib

9 лет назад, # |

-15

exactly

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AminAnvari

9 лет назад, # |

Does it have pretests or not ?

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9 лет назад, # ^ |

+47

Yeah, it will be a common rated CF round, but with problems shuffled and smooth dynamic scoring... and it will last for 2:30... and eight problems...)

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Bredor

9 лет назад, # |

← Rev. 2 →

+247

Ради майки с таким царским ником можно и написать... Как думаете, протрезвею до завтра?

А еще кстати авку было бы круто напечатать, ну так

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spOoky

9 лет назад, # |

Maxim is not orange anymore!

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9 лет назад, # ^ |

That's probably because this blog was written a few days ago as a draft when he was orange. For example for round #299, I wrote the announcement blog as a draft when I was red and I remembered to fix it just when I was publishing it.

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9 лет назад, # ^ |

so the draft was made before round #300 (i.e before about 40 days) , why would one write the announcement before 40 days of publishing it?

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9 лет назад, # ^ |

← Rev. 2 →

Well, the announcement that the round was going to be held was made around 3 months ago so it is plausible that the draft for this post was written long ago too.

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9 лет назад, # ^ |

that blog wasn't an announcement , it was the reward of those who donate to CF with at least 450$ (publishing a blog on the main page)

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9 лет назад, # ^ |

+38

Or, Monyura did this to make this blog prettier and only use orange and red colors in it.

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9 лет назад, # ^ |

+10

I agree , the list of coders looks really pretty :)

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9 лет назад, # ^ |

But Rubanenko was not red for quite some time, so it is in fact pretty old :P.

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Gassa

9 лет назад, # ^ |

Perhaps that part was just copied and pasted from the original announcement, with the dates when they had such colors, and then edited a bit.

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9 лет назад, # ^ |

+27

Didn't think I'll have to say it again :)

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9 лет назад, # |

← Rev. 2 →

+96

Btw — 8 tasks and 2h 30 min :o? Overwhelming a bit. Either most of them will be very easy or people will not solve too many of them :P.

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9 лет назад, # ^ |

← Rev. 2 →

-8

Just like round 300 and people managed to solve a lot of problems in it. Since it has problems for both divisions you need 7-8 problems to get a good spread.

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9 лет назад, # ^ |

... which definitely was not properly balanced...

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9 лет назад, # ^ |

-8

Why? As long as you solve the first three problems in less than 30 minutes (most did) you got more time for the last five than in a normal div 1 round.

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9 лет назад, # ^ |

← Rev. 2 →

+16

That's not what "balanced" means. The gap in number of solutions between F and G was huge in comparison to all the remaining gaps between successive problems. That means clogging up of the many people who solved A-F.

When there's a huge jump in difficulties, more time doesn't help at all. It just means much more people will solve F, while not so many more will also solve G.

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9 лет назад, # ^ |

-8

How would removing problems help with that?

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9 лет назад, # ^ |

"... which definitely was not properly balanced..."

"Why?"

Well I explained why that round wasn't properly balanced. I guess the point is that many problems aren't enough to get a good spread, you need problems with well-spread difficulties to get a good spread.

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9 лет назад, # ^ |

-8

Ah well, but 8 problems isn't too much for 2.5 hours at least since it would have been fine if G were like a normal D instead of a normal E.

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9 лет назад, # ^ |

See, there were no problem at all with this contest, most reds solved between 5-7 problems and the balance weren't poor either.

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9 лет назад, # ^ |

-7

"balance weren't poor either", hahahaha, very funny xD. All problem beside E and F were really simple. They took me all <=16 mins, while I couldn't get AC on F for 1,5h >_>. There was no typical "C" problem.

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9 лет назад, # ^ |

← Rev. 2 →

Even F wasn't that difficult, though it was one of those problems in which if you approached it the wrong way you'd have a lot of trouble, so it fit as the "slightly harder" problem.

(Hint: it's easy to count how many valid (l, r) intervals cross a certain point M)

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9 лет назад, # ^ |

I know how to solve it, but couldn't code it properly :/.

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9 лет назад, # ^ |

You can't really call B and C easy...

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9 лет назад, # ^ |

B is piece of cake , while you find an index i such that the guess is true then invite i.

the same problem already exist on codeforces but I'm lazy to search for it

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9 лет назад, # ^ |

B is piece of cake

Yet it took you over two hours to get it...

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9 лет назад, # ^ |

I didn't read B in early time because of the low number of who solved B. but when I did I solved it immediately

the constrains which guarantee that f[i][i]=1 for all i is a very big hint

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9 лет назад, # ^ |

Just look at how many solved it, just because it is easy to code doesn't mean that the problem is easy.

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9 лет назад, # ^ |

I didn't say easy to code, I said easy idea and that hint led me to that idea

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9 лет назад, # ^ |

BTW, here is the problem link, B already exist on codeforces as I told you.

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9 лет назад, # ^ |

So the point is that it is easy since you have seen it before?

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9 лет назад, # ^ |

it's easy because it's easy , I only remembered that it already exist later after I solved it

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P___

9 лет назад, # ^ |

Can you provide a link to the problem in contest page? Problemset is currently blocked.

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9 лет назад, # ^ |

I didn't think the block will continue for that long , here is another link anyway

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9 лет назад, # ^ |

How can you prove its correctness? What if you invite all of them, but still one of the guesses is still true? What is the correct order of processing the guests?

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9 лет назад, # ^ |

you only invite person i whenever the guess him is currently correct so when you invite person i the number of messages he will receive will be more by one than the guess so it will never happen again that the guess about him is correct.

so if I had to invite all of them then I am sure that the number of messages each one of them have got is bigger than the guesses.

also the order or inviting is not important

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9 лет назад, # ^ |

For any set of invited people, among all people where Igor guessed correctly, call the one with the lowest ID as guilty.

Now, start with the empty set of invited people; call this an invitation set. As long as there is a guilty person, invite that person as well, and call the resulting set of invited people as another invitation set. If there is nobody that's guilty, invite a random person. We keep doing this until all people are invited. This forms several invitation sets; to be precise, there are n + 1 of them, one of each of size 0, 1, 2, ..., n. We claim that at least one of these possible sets are free of guilty people.

Note that each person A can only be guilty in one set; at the first moment A is guilty, A is immediately invited into the party, and thus the number of messages that A receives exceeds Igor's prediction. Since this number will never decrease (we never reject an invited person), A will always get more messages than Igor's prediction afterwards, so A will never be guilty any more.

However, there are n + 1 invitation sets in total, but only n people that can be guilty. By (generalized?) Pigeonhole Principle, one of the sets will be free of guilty people. This is the set we want.

To make it constructive, instead of picking a random person when there's no guilty people, stop and take that set instead; we are guaranteed to stop for the same reason as above. This is the basis of my submission.

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9 лет назад, # |

combined division again, what are the chances to be one of the top 200 :P

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9 лет назад, # ^ |

← Rev. 2 →

I'm more interested in how high the chances to get +200 are. Also chances to get +400 and +600. With this number of participants and a combined round...

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9 лет назад, # ^ |

← Rev. 2 →

+52

We have 737 red coders now, let's wait and see how will this number change tomorrow :)

P.S. And also 46 people with IGM, let's also check this one after round.

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9 лет назад, # ^ |

Opps, Red Alert!!

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9 лет назад, # ^ |

743/48, not so bad. Did they finally resolved issues with inflation? :)

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etal

9 лет назад, # ^ |

Oh, you made my day! I thought I had become red (for the first time) just because of rating inflation :')

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9 лет назад, # |

+13

Wow, eight problems! I think many hacks will be in this contest.

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BBM

9 лет назад, # ^ |

unfortunately your comment has been hacked :D

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9 лет назад, # ^ |

+11

If I stay in the same room with you today, I will hack you :D

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KungA

9 лет назад, # |

+93

Кажется случайный порядок задач вносит погрешности в результаты. Это нормально при команде из 3-х человек и 15 часах времени (на команду), и сомнительно при 8 задачах на 2.5 часа (+ падение стоимости, а не штрафное время).

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Lo_R_D

9 лет назад, # ^ |

-74

Да, 15 часов на команду обычно хватает, чтобы решить всё... а еще, думаю, хватит времени поспать, поесть два раза и опоздать часов на 10 к началу соревнования (в случае Гены — можно опоздать на 12 часов).

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Lo_R_D

9 лет назад, # ^ |

-63

Еще ты забыл про взломы, благодаря которым, даже в случае одного успеха, можно не хило вырваться вперёд.

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nk.karpov

9 лет назад, # ^ |

+21

Если задачи короткие, то норм :)

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ThunderStroke

9 лет назад, # |

← Rev. 2 →

I love green and also the codeforces logo :)

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9 лет назад, # |

← Rev. 3 →

+219

By the way, I have a feeling that this needs to be explicitly expressed by someone — I think that this contest has the best prizes on CF ever! There are 200 (!!) T-shirts and they are personally signed (!!). I don't know what these Oculus Development Kit do (and don't have time to delve into it), but they look cool and valuable. And these thingies to people from places 16-50 looks completely useless but amazingly funny, which makes me consider whether I should do some unsuccessful hacks if I will see that I have chances for being in TOP15 :3

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9 лет назад, # ^ |

+127

"makes me consider whether I should do some unsuccessful hacks if I will see that I have chances for being in TOP15 :3"

#justgrandmasterthings

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Martynas

9 лет назад, # ^ |

+27

Thanks for the clarification Swistakk – before reading your comment I thought that "competitors from 16 to 50" means people aged between 16 and 50, which raised a lot of questions :)

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9 лет назад, # ^ |

-16

makes me consider whether I should do some unsuccessful hacks if I will see that I have chances for being in TOP15 :3

Consider this case: You are in top 15, you attempt some unsuccessful hacks on purpose to be in 16-50. In system testing one of your solution fails and you are not in top 50 anymore. Then you notice without that unsuccessful hacks you would be in top 50.

This case is very sad. And if something like this ever happens to me, I'll say goodbye to competitive programing for good.

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9 лет назад, # ^ |

Yeah, that was rather just joking :P. Btw saying goodbye to CP because of such a reason would be pretty stupid xD.

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9 лет назад, # ^ |

If such case happens, CP would say good bye to you

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WYJOIER

9 лет назад, # |

-27

Can I ask how many problem in this contest ?

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Tung_Duong

9 лет назад, # ^ |

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WYJOIER

9 лет назад, # ^ |

Thank you ^^

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SomeRandomGuy

9 лет назад, # |

← Rev. 2 →

+28

340 red coders registered to contest. No chance for experts

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alibi

9 лет назад, # ^ |

← Rev. 2 →

+78

Never give up! Impossible is possible!!

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9 лет назад, # ^ |

If someone can win, then you can also win. Always be optimistic.

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t3rminated

9 лет назад, # ^ |

there was never

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9 лет назад, # ^ |

+10

see there is an expert under 200 rank in final standings :)

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jason_yu

9 лет назад, # |

The time is rarely suitable for coders in China. :) PS. The College Entrance Examination will start tomorrow morning. Good luck! God bless us!

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SomeRandomGuy

9 лет назад, # |

-8

Rubanenko почему-то красный в анонсе.

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9 лет назад, # ^ |

+26

Это сделано из уважения к его вкладу в подготовку контеста :)

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9 лет назад, # ^ |

+138

У меня это в контракте прописано.

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9 лет назад, # |

Today's contest is a big opportunity for Div2 participants to become members of Div1.

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Michael_Carrick

9 лет назад, # ^ |

how ?

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Cactus_Irl

9 лет назад, # ^ |

If you do better than Div1 contestants, you'll get a lot of rating.

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nezametdinov

9 лет назад, # |

+16

Izi problem, izi life

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Zharaskhan

9 лет назад, # ^ |

+13

Izi futbolka, izi life

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CMaster

9 лет назад, # ^ |

Hard problem :) hard life :)

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9 лет назад, # |

← Rev. 4 →

So great t-shirts *-*

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9 лет назад, # |

-10

expecting unusual rating changes in this contest like: Specialist -> Candidate Master; Expert -> Master ; Candidate Master -> International Master, etc.

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9 лет назад, # ^ |

And suddenly International Master -> Specialist. The most unusual, I think.

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9 лет назад, # ^ |

Generally in div 1 + div 2 contests maximum +ve rating change is around +400 but maximum -ve rating change is only around -135. So I don't think you are correct. :P

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I_love_Hoang_Yen

9 лет назад, # |

+10

I'm going to be on airplane during contest time..

But airplane has (unstable) wifi..

Not sure if I should participate..

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9 лет назад, # ^ |

You should try!

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Reyna

9 лет назад, # ^ |

-13

You should catch!

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9 лет назад, # ^ |

Create fake account xD

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I_love_Hoang_Yen

9 лет назад, # ^ |

-12

Well sure I could do that, but then how to avoid following conversation:

GF: Wow, you won Looksery T-shirt
Me: (put some cute sentence here)
GF: Wait, where is "I love Hoang Yen"? :-O

(of course there's option like creating account I_love_Hoang_Yen_2, but it would be too obvious :)))

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9 лет назад, # ^ |

+36

You could have created an account "I_love_cutest_girl_in_world" and tell her that it is obvious that is about her, nobody will have strong suspicions about you and that T-shirt would be more 'universal' :pp. That means that it won't become out-of-date whatever the future will be xD.

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I_love_Hoang_Yen

9 лет назад, # ^ |

+17

Do you think that such lame trick can not be seen through by the eye of a girl? :p

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Resul

9 лет назад, # ^ |

← Rev. 2 →

I_love_you written t-shirt would be nice present to your girlfriend...

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9 лет назад, # |

+88

tourist began preparing to travel to San Fransisko. ;)

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Resul

9 лет назад, # ^ |

+50

With his Tourist written t-shirt. He is really tourist in San-Francisco xD

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-Wave-

9 лет назад, # |

The worst thing about random order is trying to find the "A" problem at the very beginning. But today I have a secret weapon: https://www.random.org/

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9 лет назад, # ^ |

+18

Now it's not secret anymore :P

seriously speaking, I will tell you my strategy in such situation you might like it:

I open all the problems, then I read shortest problem statement first then second-shortest and so on until 10 minutes after the starting time I open the dashboard to see if some problems are solved if yes I read them then I solve them until I finish solving them ,the difficulty of the problems will be revealed.

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sepehr

9 лет назад, # |

that will have random order at the contest

Painful :((

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ThunderStroke

9 лет назад, # |

I think this time tourist can't make him first :/

may be other one will be the first ;)

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9 лет назад, # |

← Rev. 2 →

-11

scoring dynamic? ):

random problems? ):

I think that Problem D is the easiest problem in this contest (:

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9 лет назад, # ^ |

And that's because... ?

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9 лет назад, # ^ |

because I am lucky (:

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9 лет назад, # ^ |

+51

Oh, don't make us to reshuffle them again!

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9 лет назад, # ^ |

Ok (: I'm sorry

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ecapia

9 лет назад, # |

Hello everyone !!

I hope high rating and good luck for everyone.

I have a question .

I have heard a lot of people said when the two division participate together is easier will have a high rating.

Why ??? Anyway I thank for you answer in advance!!

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9 лет назад, # ^ |

-8

We simply want more people to have high rating ;)

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monkey.coder

9 лет назад, # ^ |

2 reason i think (at least for div2 participant) : 1. less "strong" unrated coder will participate (iykwim) 2. div2 participant can defeat div1 participant which can gain more rating

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lightstone

9 лет назад, # |

← Rev. 2 →

+11

Does random order mean that it is shuffled by a computer and is completely random or are they put in an order developers want them to be?

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9 лет назад, # ^ |

+22

Sometimes developers' decisions are more random than the computer's ones

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dened

9 лет назад, # |

+12

Когда уже другие элементы одежды будут в качестве призов довать? А то у меня маек уже хватает, а вот шорт, например, нет. :)

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imcoolkanaut

9 лет назад, # |

← Rev. 3 →

some random t-shirts for us peasants would be much appreciated :)

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9 лет назад, # |

finally 5808 registrants :)

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AkshajK

9 лет назад, # |

Can you delay contest 5 min to allow for more registrants?

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nirob_mh

9 лет назад, # |

happy coding :)

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boleyn.su

9 лет назад, # |

+13

I love the prizes! GL & HF to everyone!

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soshika

9 лет назад, # |

hey guys look at unknownuser96 in room 118 link

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andreyv

9 лет назад, # |

+46

(╯°□°)╯︵ ┻━┻

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Irkutsk_ISRU

9 лет назад, # |

+10

Задачи очень интересные! Спасибо за контест! :)

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9 лет назад, # |

D problem author,great problem, like!

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Kvark161

9 лет назад, # |

На чем С ломали?

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taxicoo

9 лет назад, # ^ |

У многих не заходило 2 2 1 2

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passick

9 лет назад, # ^ |

Подтверждаю, пятерых так взломал.

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Omelianenko

9 лет назад, # ^ |

5 3 1 1 2 2 1

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halin.george

9 лет назад, # ^ |

На случае, когда K == N, мне кажется. Во-первых, потому что меня так взломали, во-вторых, этот случай легко можно было разобрать, смотря в код человека, который взламываешь. Смотришь на исход программы взламываемого, и там уже видно.

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Fear_Is_An_Illusion

9 лет назад, # |

Hard would be an understatement , at least for poor coders like me

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9 лет назад, # |

← Rev. 2 →

How is C solved? I could only think of a O(N²) DP solution with O(N) space complexity.

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VeniVidiVici

9 лет назад, # ^ |

A lot of casework. Whoever goes last will generally win, unless it is possible for Daenerys to burn all the towns with odd number of residents, it is possible for Daenerys to burn all the towns with even number of residents and k is even, or it is possible for Stannis to burn all the towns with even number of residents and k is odd.

Unfortunately, there is also the sad final case of n=k, in which case nobody gets a turn and so there is no "going last"; in this case answer simply depends on number of towns with odd number of residents.

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9 лет назад, # ^ |

← Rev. 2 →

+12

Do it in O(1). There's a categorization (although not really simple) based on the number of cities with odd/even populations (called odd, even respectively), something like this (not sure, though):

If n = k, the game has ended, just count the number of cities with odd populations.
If $\text{[math]}$ , Daenerys always wins; Daenerys always bombs an odd city.
If $\text{[math]}$ (but not the above), the winner depends on $\text{[math]}$ (it's Stannis if odd, Daenerys if even); the winner always bombs an even city.
If none of the above applies, the winner depends on $\text{[math]}$ (that is, the number of moves; again, Stannis if odd, Daenerys if even); the winner plays the city opposing the previous move, to guarantee that he has a choice for the final move.

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-XraY-

9 лет назад, # ^ |

+62

I hate the case n = k. Found it as challenging, killed two solutions, but mine is still wrong. Stupid problem, a lot of points and much more stupid bug. :(

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9 лет назад, # ^ |

+24

Unfortunately, there is also the sad final case of n=k

"Sad" is very polite word for this case :)

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9 лет назад, # ^ |

Would you please explain your third case?

How does it is depending on the pairity of k?

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9 лет назад, # ^ |

If all even cities can be destroyed by one player, then all remaining cities are odd; there are exactly k of them. By checking the parity of k, we can check which player should destroy the even cities; if k is odd, then Stannis should destroy them (and Daenerys can't do anything to prevent that), while if k is even, it's the other way around.

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Ripatti

9 лет назад, # ^ |

← Rev. 2 →

+13

Just coded stupid O(n²) solution to see patterns and then checked clever O(1) solution by the stupid one.

upd. Too bad, it failed. But I think the right approach should be like this.

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xuanquang1999

9 лет назад, # ^ |

← Rev. 2 →

Call the number of odd houses is odd, and the number of even houses is even

If Stannis can burn all the even houses before the game end, he can force Daenerys to burn only the odd houses in later move (after all even houses run out), and if the number of odd houses after the game is odd, he will win. The condition to do this is (n-k+1) div 2 >= even and odd-(n-k-even) mod 2 = 1.

If Daenerys can burn all the odd houses before the game end, he will always win. The condition to do this is (n-k) div 2 >= odd.

In all other case, the person who end the game is winner. Why? Because in all move (except the final), the players can make opposite move to each other (if Stannis burn odd house, Daenerys will burn even house and vice-versa). And the player who do the last move can just perform the move that give him victory.

→ Ответить

9 лет назад, # ^ |

← Rev. 2 →

+16

I solved that with DP , state is player , CitiesToBeDestroyed , RemainingOddCities

Note that RemainingEvenCities can be omitted since it can be drived from the other parameters .

Till now it is O(N*N) , since both players will play optimally and each one of them will try to cancel each other move , I removed x odd city & x even city & 2 * x CitiesToBeDestroyed before calling the DP ( left at most 128 for the DP to figure out a solution )

Thus it will be O(N * 128)

Take a look at my solution 11465922

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9 лет назад, # ^ |

BTW , I got WA on pretest because of a logical mistake , so I increased the factor when I was debugging . 128 is too much (8 is enough to get AC) ==> O(N*8)

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thecodekaiser

9 лет назад, # ^ |

Can you please explain more on how you came to the intuition that you can remove x odd cities and x even cities before calling the DP(leaving at most 128 for the DP to figure out a solution) ?

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Scorpy

9 лет назад, # |

KareemMohsen and MostafaG submissions for last round look suspicious.

11467747 = 11467830 11461749 = 11467680

P.S. It's impossible to view these submissions until the end of contest, but I was in the same room with them and

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9 лет назад, # |

Anybody solved D with fenwick tree ? I couldn't.(

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allllekssssa

9 лет назад, # ^ |

+16

I don't understand problem D ! I work my tasks about 1 h ( first 10 minutes and last 50 ), all other time I spent for reading D.

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9 лет назад, # ^ |

Optimal solution is O(n * m)

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9 лет назад, # ^ |

I will cry now (. What was the solution for D?

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9 лет назад, # ^ |

← Rev. 2 →

Let b[x][y] be 1 if (x,y) is 'W', -1 if it is 'B', 0 if out of bounds.

Increase answer by one for every (x, y) in board such that b[x][y] != b[x+1][y] + b[x][y+1] — b[x+1][y+1].

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Hyperbolic

9 лет назад, # ^ |

why? please explain more ...

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9 лет назад, # ^ |

+11

Every rectangle with bottom-right corner (x,y) is added a number of times R_x, y to the final answer (it might be 0 if we never use it).

Because you're only doing linear operations, the final answer for any sum of rectangles is C_x, y * a_x, y, where a_x, y is the value on the board at position (x,y) and C_x, y is the sum of R_x, y for all rectangles that contain (x,y).

But how can we calculate C_x, y easily? Well, except for the rectangle with bottom-right corner (x,y), all rectangles that influenced (x,y) are the ones that influenced either (x+1,y) or (x,y+1). To eliminate double-counting, remove the ones that influenced both (x+1,y) and (y,x+1): the ones that influenced (x+1,y+1). Our final result is: C_x, y = R_x, y + C_{x + 1, y} + C_{x, y + 1} - C_{x + 1, y + 1}.

To get the sum we're looking for, we want final C_x, y to be equal to b_x, y for every (x,y), so from the earlier equation R_x, y = - b_x, y + b_{x + 1, y} + b_{x, y + 1} - b_{x + 1, y + 1}.

After you calculate all R_x, y, count the ones that are not zero, and that's your answer.

→ Ответить

9 лет назад, # ^ |

Is this based some kind of standard method? I'm trying to understand why this problem was solved by so many.

→ Ответить

stargazer

9 лет назад, # ^ |

O(n^2*m^2) algorithm was enough for this problem.

→ Ответить

satyaki3794

9 лет назад, # ^ |

← Rev. 3 →

I used a similar method. For cells not lying on the right or bottom border, I considered all cases in which that prefix rectangle would be used atleast once (how many times, does not matter as we can just multiply the required integer, after using it once). It will always be used unless its right, bottom and bottom-right cells are similar to it or exactly two of them are different and are adjacent.

For the bordering cells (right or bottom border) they would be used if they are not equal to adjacent cell on the border.

Please look at my submission for a better understanding. Solution complexity if O(n*m).

11473484

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9 лет назад, # ^ |

My solution is O(N²M²). I think 2D Fenwick Tree is unnecessary since N² * M² ≤ 10⁸

→ Ответить

9 лет назад, # ^ |

Please explain your approach.

→ Ответить

9 лет назад, # ^ |

← Rev. 2 →

The same here.

Init sum for every pos as 0.

Iterate for every cell from bottom-right corner to top left corner.

If our mask at this pos == 'W' && sum for this pos != 1:

sum of every pos from this to top left corner += 1 — d[i, j]

result++

If our mask at this pos == 'B' && sum for this pos != -1:

sum of every pos from this to top left corner += — 1 — d[i, j]

result++

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9 лет назад, # ^ |

i was unsure whether O(n^2*m^2) would pass. hence i implemented an O(n^2*m) solution.

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9 лет назад, # |

← Rev. 2 →

Why O(255025000000) operation is not TL ? 11473557

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mkirsche

9 лет назад, # |

← Rev. 3 →

+24

Was Problem E just this paper? Here

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ifsmirnov

9 лет назад, # ^ |

Your link seems to be broken.

→ Ответить

http://theory.stanford.edu/~megiddo/pdf/circspr.pdf

9 лет назад, # ^ |

→ Ответить

mkirsche

9 лет назад, # ^ |

Oops, it converted one of the backslashes into 2 backslashes. Fixed.

→ Ответить

9 лет назад, # ^ |

here

→ Ответить

Lewin

9 лет назад, # ^ |

← Rev. 2 →

+29

That paper seems much stronger than what we need. Also, it seems it may be numerically instable, and also seems a bit hard to implement.

I believe there's a way to do this using only integers, though the implementation is still very tricky since there a lot of special cases. My approach was to make the points (x,y,x*x+y*y), then find whether or not these two sets can be separated by a plane. This involves checking whether or not the two 3d convex hulls intersect. The implementation is made slightly simpler by allowing quadratic solutions. Unfortunately, there are a lot of degenerate cases when things are coplanar, and I wasn't able to deal with all of those in time.

EDIT: Actually, the paper seems to talk about my approach in section 4. Should have read it a bit more carefully.

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__builtin__wolfy

9 лет назад, # ^ |

I tried to compute the smallest enclosing circle for both sets and check if the other points lie outside this circle but I got wrong on #10. Is this even correct?

→ Ответить

anzu

9 лет назад, # |

Interesting problems . Thanks for good contest .

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RetiredAmrMahmoud

9 лет назад, # |

+150

I understand you want to introduce your company's work in the statements. But WTF was problem D's statement?

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ecapia

9 лет назад, # ^ |

← Rev. 2 →

+31

I agree with you, THe problem D was very difficult understand I needed more than 45 min in order to understand MY english is bad I think that was that. but is the first time that happened this. did Someone have the same problem??

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9 лет назад, # ^ |

English does not matter, I understood the problem and sample test cases after reading for 30 mins, yet I could not solve it

→ Ответить

9 лет назад, # ^ |

← Rev. 2 →

I spent 30 min trying to understand the statement and the result I didn't , spent 2 mins to understand the problem from sample explanation .

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PavelKunyavskiy

9 лет назад, # |

+39

Классные задачки! Хотя, кроме F и E все простые, над всеми пришлось подумать.

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pi37

9 лет назад, # |

Hoping for a systest miracle: for enough people to fail for 278th to get to top 200

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accidentallygivenfuck

9 лет назад, # |

+117

Winner is from North Korea. Hmmm... Congratulations to vepifanov with a trip to San-Francisco. :P

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jaguar1996

9 лет назад, # ^ |

winner is tourist :D

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9 лет назад, # ^ |

sure

→ Ответить

ilyakor

9 лет назад, # |

+49

How to solve H? I've spent literally the whole contest on it :( Still getting WA#4, and I have no idea why.

→ Ответить

cerealguy

9 лет назад, # ^ |

+11

overflow

→ Ответить

9 лет назад, # ^ |

← Rev. 2 →

+17

Binary search; when you have fixed some answer, you have intervals of possible values for a,b,c,d in matrix B. From those invervals you can get invervals for values of ad and bc. Now you only need to check them for intersection.

→ Ответить

9 лет назад, # ^ |

Now you only need to check them for intersection

Would you please explain it?

In this problem, how can I tell from the values of ad and cd that our current answer is valid? I don't understand it.

→ Ответить

9 лет назад, # ^ |

← Rev. 2 →

Assuming answer is equal to X, you can pick any value in range [A-X,A+X] (I denote values of original matrix by A,B,C,D here) for a and any value in range [D-X,D+X] for d. This gives you some range [L1,R1] for possible values of product of a and d. By picking proper values for a and d in your new matrix you can reach any resulting product from that range. In the same way you can get range [L2,R2] of possible values for product of b and c. You are interested in picking such values for a,b,c,d that ac-bd=0, therefore ac=bd, which implies that [L1,R1] and [L2,R2] have a common point. I hope my code (11475874) is clear enough.

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9 лет назад, # ^ |

Yep got that. I was trying to make the matrix B. That was my mistake.

Thanks.

→ Ответить

-Wave-

9 лет назад, # ^ |

Yup, same situation. I figured out how to solve it if (a+b+c+d)=/=0, but then got stuck. Really frustrating when you see the number of people that solved it...

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9 лет назад, # ^ |

I feel your pain... Spent whole contest on H, in the end I ended submitting random calculations until this passed pretests, but I have absolutely no idea why:

for (int bs = -1; bs <= 1; bs += 2) {
		for (int cs = -1; cs <= 1; cs += 2) {
			for (int ds = -1; ds <= 1; ds += 2) {
				if (bs*cs == ds) {
					long long den = -a + bs*b + cs*c - ds*d;
					if (den != 0) k = min(k, fabs((a*d - b*c) / (double) den));
				}

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Muh

9 лет назад, # ^ |

← Rev. 2 →

I can't understand why this works. My solution had 8 cases:
1) (a+x)(d+x)=(b+x)(c+x);
2) (a+x)(d+x)=(b-x)(c-x);
3) (a+x)(d-x)=(b+x)(c-x);
4) (a+x)(d-x)=(b-x)(c+x);
5) (a+x)(d-x)=(b+x)(c+x);
6) (a-x)(d+x)=(b+x)(c+x);
7) (a+x)(d-x)=(b-x)(c-x);
8) (a-x)(d+x)=(b-x)(c-x);
This solution don't work. But when i sent just first 4 cases have AC.
Why?
Example, 5th case: ad-bc=2*sqr(x)+x(b+c+a-d), when discriminant>=0 we can find new answer. If I am wrong please explain, all the time I spent on this task.
UPD when Discriminant > 4*(10e18) my solution broked.

→ Ответить

vippu95

9 лет назад, # ^ |

← Rev. 2 →

My solution worked with above 8 cases.. I was trying to do the same during contest..but forgot to add the case of 0. After contest i added the case to handle 0 and got accepted.

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poikniok

9 лет назад, # ^ |

I am not positive this works but basically binary search. The truth condition is if by changing each of the elements by this much you can get the opposite sign of the determinant. Maybe this fails somewhere but I don't think so.

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poikniok

9 лет назад, # ^ |

Idea was correct, but WA50 because of precision issues I think. Rewrote it with BigDecimal and now it works. Lesson learned I guess.

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9 лет назад, # ^ |

Me too. ;-(

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9 лет назад, # ^ |

I suspect that there might actually be a simpler solution to this. Note that in the samples, for B each element differs by the same amount from its corresponding element in A (that is, |1.2 - 1| = |1.8 - 2| = |2.8 - 3| = |4.2 - 4|). I tried proving it, although I don't think it was rigorous or even worked. However, if it holds, then it's just checking 2⁴ cases: add or subtract a number t to each element in A (that's where 2⁴ comes from), compute its determinant (a quadratic expression in t), and find its solutions. Take the one with minimum magnitude among all found solutions.

→ Ответить

bluepichu

9 лет назад, # ^ |

I can't prove it, but I can say that (with this testset) it worked, since that's exactly the approach I took. See 11469632.

→ Ответить

9 лет назад, # ^ |

← Rev. 2 →

Yes, I also used the same approach and passed.

The idea is that if we fix the top row of the matrix B to be x, y, we know the bottom row is in the form xt, yt for some t; since the graph of $\text{[math]}$ is a polyline (several line segments/rays joined together), the minimum must be found at an intersection, where |c - xt| = |d - yt|. Generalizing that this must happen to all the elements at the same time is the shaky part.

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jibon_ebong_shomoy

9 лет назад, # ^ |

← Rev. 2 →

Did it work?

→ Ответить

9 лет назад, # ^ |

Check my O(1) solution for H. I just tried all the possible combinations of +x or -x at all the places of the matrix. SOLUTION

→ Ответить

deepak2015

9 лет назад, # ^ |

Can u plz explain ur solution in detail. Sorry for inconvenience.

→ Ответить

sonjoydabnath

9 лет назад, # ^ |

explain plz?

→ Ответить

EvgeniSergeev

9 лет назад, # ^ |

For the record, here is another constant-time solution for H, except that it relies on arbitrary-precision arithmetic to get the accuracy to 1e-9: 11493234

The explanation is based on similar thinking to the editorial, treating the matrix as two vectors. After drawing some pictures, we need to find a line passing through the origin (i.e. this is the degenerate matrix B) such that the distances along diagonal lines from the the vectors to this line are the smallest possible (i.e. the largest of them is minimised). Diagonals are usefully dealt with by rotating 1/8th turn, then it becomes a matter of dropping down vertical lines from the vectors onto a line of the form y = mx. And the rest are details.

But I'd really like to know if there is a simple reason why prateek_jhunjhunwala's solution works, or is it just a coincidence?

→ Ответить

9 лет назад, # |

← Rev. 2 →

How to solve problem D? Is it possible to solve O(n*m) is the optimal solution?

→ Ответить

sonjoydabnath

9 лет назад, # ^ |

← Rev. 2 →

Yep it's possible to solve in O(n*m). Though I 've solved it after Contest. my O(n*m) solution 11478809 .

you just do not need to update the whole prefix rectangle. :)

→ Ответить

sonjoydabnath

9 лет назад, # ^ |

How did you came up with O(1) solution of pH ? will explain it for me ?

→ Ответить

9 лет назад, # ^ |

← Rev. 4 →

+27

→ Ответить

9 лет назад, # ^ |

← Rev. 7 →

or here You can also see my code 11506983

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9 лет назад, # |

Congratulations I_love_you ! will be nice t-shirt with this handle ;)

→ Ответить

Hyperbolic

9 лет назад, # |

Test is ended. Please give me good score~~~

→ Ответить

ifsmirnov

9 лет назад, # |

+11

Задачки идейные и в целом хорошие, спасибо. По много минут думать над G и H, чтобы потом за две минуты закодить -- приятно.

Кажется, правда, между E,F и остальными задачами получился большой разрыв по сложности, о чём ещё перед контестом разные комментаторы писали.

→ Ответить

Stochastic

9 лет назад, # |

+82

Обменялись любезностями.)

→ Ответить

chEEtah

9 лет назад, # ^ |

+27

=)) Да, забавно получилось. Я увидел, что меня взломали, и пошел мстить)

→ Ответить

cerealguy

9 лет назад, # |

+90

242D - Dispute

→ Ответить

ecapia

9 лет назад, # |

← Rev. 2 →

+39

THe problem D was very difficult understand I needed more than 45 min in order to understand MY english is bad I think that was that. but is the first time that happened this. did Someone have the same problem??

→ Ответить

votick

9 лет назад, # ^ |

I finished problem A then spent the rest of the contest trying to understand D, I jumped to some other problems as well but since so many people were finishing D I kept going back to it.

Still don't know what it was asking for O_o

→ Ответить

allllekssssa

9 лет назад, # ^ |

Now, I don't understand problem :)

→ Ответить

9 лет назад, # |

+166

Seems like RNS hadn't even tried to pretend to be single person. 2 submits in the very same second, 2 accepted within 14 seconds of each other...

→ Ответить

Krasnokutskiy

9 лет назад, # ^ |

+13

Also the two neighbours submissions have pretty different amount of includes...

→ Ответить

RetiredAmrMahmoud

9 лет назад, # ^ |

+20

I think I've read somewhere before that they are a team from Korea always participating together.

I wonder what they are going to do with the tshirt. smh

→ Ответить

9 лет назад, # ^ |

+22

I do believe in integrity of Codeforces contests enough to be sure there would be no T-shirt for them

→ Ответить

9 лет назад, # ^ |

+122

We're aware of this "user", this won't be left unpunished.

→ Ответить

9 лет назад, # ^ |

And his name is nowhere in the rank list. Faith in codeforces restored (not that it was ever shaken)!!

→ Ответить

9 лет назад, # ^ |

it's because two of their solutions failed system test

→ Ответить

9 лет назад, # ^ |

+24

I feel like having a big banner on CF home page saying "NORTH KOREA DOES IT AGAIN: ANOTHER SHAMEFUL CHEATING ATTEMPT FROM NORTH KOREAN STUDENTS" would be a more appropriate punishment than hiding them from the leaderboard.

→ Ответить

9 лет назад, # ^ |

← Rev. 3 →

+41

I think that that kind of generalization would be too racist.

→ Ответить

9 лет назад, # ^ |

+10

Suppose Brazil U20 wins the World Cup 3 times in a row. The day after the 3rd final you open the newspaper and it says "Brazil does it again: another amazing performance from Brazilian youth". Does it sound a racist generaliazation?

→ Ответить

9 лет назад, # ^ |

You can't say it's shame on all north korea coders just because one of them cheated

→ Ответить

9 лет назад, # ^ |

+13

It was definitely more than 1 though.

→ Ответить

9 лет назад, # ^ |

← Rev. 4 →

Your example is just silly. It doesn't sound like a racist generalization because it isn't. One might claim that usage of "Brazil" and "Brazilian youth" is a generalization since it's actually only a team who've won the cup, not the whole country/nation. But it's wrong because that is their national team we're talking about. That sentence congratulates the whole country because the team that represents them have won. It's totally different than blaming a whole nation based on limited observation. I think you should consider what you've said before and try to be less offending.

→ Ответить

9 лет назад, # ^ |

I do not understand your argument. You were happy with Brazilian headline. Now I went ahead and replaced Brazil with NK, the phrasing referring to the act of stellar performance with the phrasing referring to the act of shameful cheating, and Brazilian youth, that represented the team, with NK students, that represented the NK lab behind the user.

→ Ответить

9 лет назад, # ^ |

I still think that my argument is clear, but I will try to put it in other words just to be sure:

There are 10 people in a room.

First case: Some person steals something from me. Then, I come to conclusion of every person in that room is a thief.

Second case: I want someone to play ping pong with me, that's why I go into that room and ask them to chose a representative player for their room. They choose the player. I play with that player. He wins. I congratulate the room.

The important thing to notice is: I'm not saying everyone in this room can beat me. I'm just saying that this room has defeated me. In the second sentence, "the room" actually refers to the player that represents its room.

First case is a generalization. Second one isn't. It's not related to action's morality or something. It's just that these examples are different. Trying to compare them is absurd.

→ Ответить

9 лет назад, # ^ |

+22

Now I realize you probably don't possess all the information. This is not the first or second case of a user claiming to be from North Korea being in the top through obvious cheating (code strongly resembling other NK user's code) or less obvious suspicions (submits for different problems within seconds of each other, different coding style, etc). Sometimes they get punished (like on Rockethon this year), sometimes they don't (there was a long discussion on Codechef sometime ago about a Long contest, not sure what happened in other Long contests). I haven't encountered a single case of NK user in the top that resembled an honest result. (I believe there are links pointing to some of these occasions in comments around here so you can check for yourself.)

Hence analogy to your example would be -- every time I see a user from that room coming out, he is stealing something. And I come to conclusion that that room is full of thieves.

→ Ответить

9 лет назад, # ^ |

Hiding them from leaderboard doesn't prevent us from something like you've described. The cheating issue is much more general and complex than just punishing separate users that get too high in the leaderboard.

Also that would be not fair to say that the worst cheaters are from North Korea, there are many other countries with high amount of cheaters...

→ Ответить

9 лет назад, # ^ |

North Korean cheaters are the only ones I've seen hailed in internet media (I am referring to NK press making news out of their 'victories' in online contests). I suspect they wouldn't continue this blatant cheating if they got publicly scolded once.

→ Ответить

9 лет назад, # ^ |

I think they wouldn't care. It's like they're cheating for the sake of cheating, being that obvious.

Not a big banner highlighting someone, but a Hall of Shame (in a sufficiently non-annoying manner) for all cheaters would be appropriate.

→ Ответить

9 лет назад, # ^ |

They were quite obvious at Codechef contests too, yet Codechef chose not to ban them. They are actually posting news about results of those contests to show how advanced their university is.

→ Ответить

9 лет назад, # ^ |

Mhm, I wonder why Codechef did what they did. My guess is negligence, but being threatened is also a possibility.

You can't stop propaganda, anyway. If they didn't win any contests, they could outright lie.

→ Ответить

adamant

9 лет назад, # ^ |

+19

I wonder if Korean students are really aware that they are participating not in a team competition?

→ Ответить

Live_Forever

9 лет назад, # ^ |

+28

Why should they ? The believe these are team contests.. :P Here Here

→ Ответить

slycelote

9 лет назад, # ^ |

-6

That's hilarious! "Chinese team ACRush working for Google in Washington"

→ Ответить

9 лет назад, # ^ |

+20

Is that all the proof you have?

I'm pretty sure I've had 2 accepts in 14 seconds at some point in some contest. You give up on debugging a problem, do something else, go back to debugging the first one and find the bug immediately.

Also, if the internet connection is down, you'd continue solving things and then submit 2-3 solved problems at the same time when you're able to submit.

→ Ответить

9 лет назад, # ^ |

+29

I do not want to go deeper into 14 seconds delta, but non trivial submits on different problems in the same second?

→ Ответить

9 лет назад, # ^ |

I explained it. Connection comes back up, you submit 2 things you solved during downtime. Some random lag causes the first submission to delay by 2-3 seconds, so it looks like the same second instead of 5 seconds difference.

→ Ответить

9 лет назад, # ^ |

+24

While using different templates for different problems

→ Ответить

9 лет назад, # ^ |

That is slightly strange, but still not proof. For example, when a problem requires maxflow, I copypaste some code that I wrote years ago, using a completely different coding style and different macros.

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9 лет назад, # ^ |

+66

There ought to be a moment when you stop multiply small probabilities and say that something that walks like a duck and quacks like a duck is a duck. Besides admins can check you theory about disconnect by looking into logs

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9 лет назад, # ^ |

Obviously it looks 99% like what you're saying, but you have to give an extreme amount of "innocent until proven guilty", since false positives are quite terrible.

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9 лет назад, # ^ |

+21

No, it looks maybe 99.9999% like what you're saying. The chances of getting 2 submits on the same second are already extremely low — it's never happened to me before, AFAIK, and I've made a LOT of submits. A few seconds are very, very rarely enough to fix a bug and submit. Especially if "you're on a slow connection".

By quantum physics, there's non-zero probability of an event occurring (as long as it's in accordance with basic laws — causality comes to mind here). How low does the chance of being a false positive have to be for you to disregard it?

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9 лет назад, # ^ |

to W4yneb0t:

I don't know why are you arguing whether it's cheating or a rare cased happened.

simply if it is really rare case happened at least RNS could say so,but even him didn't say that it's rare case happened and if he is banned from commenting then he could create new account to say so ,or he can ask a friend to use his account to comment.

also if you're saying that those proofs are only 99% correct then could you give us example of how a 100% proof look like?

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9 лет назад, # ^ |

← Rev. 2 →

As Xellos pointed out above, there's no such thing as a 100% proof. It's hard to say what's enough, but 99.999% definitely is and 90% definitely isn't. With only the info in Egor's original post (before he and Xellos responded with more info), I wouldn't even give it 90%.

This is not another matter of two gray users submitting the exact same code with 1 character changed. Take it more seriously.

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9 лет назад, # ^ |

-11

if I was in RNS situation and I know that I'm not a cheater then I would post a comment with explanation to how is that these rare cases happened with me.

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9 лет назад, # ^ |

+18

While bragging about competing as a team.

Anyway, which of these two options do you think is better?:

skip solutions of suspected cheaters and give them a chance to appeal it
nothing

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9 лет назад, # ^ |

Bragging where?

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9 лет назад, # ^ |

One link (Russian, Ctrl+F AcRush), evidence that at least one NK team considers individual participants "teams". Further links showing mugurelionut's digs into it: team rns4 (presumably this RNS) did cheat by using another account to submit.

Just try googling for more. Okay, one more link, compare it with this situation.

At this point, you're asking if it's "all the proof" that there's of NK sending teams into individual contests and publicly talking about winning them (hence "bragging"); repeatedly, NK "users" have successive submits with extremely small time gaps, different users had nearly identical codes (plus CC decided they didn't cheat, which is suspicious by itself), and individual users repeatedly send codes with completely different structures. Once again: extremely similar behaviour for different users. Yes, that's all the proof.

Anyway, you haven't answered my question: which choice is better?

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9 лет назад, # ^ |

+16

Ok, I'm largely convinced. I'm not sure which is better — some people would just say "well fuck you then" instead of appealing, and even if they appeal successfully, there will be some damaging community drama.

I suppose whoever made the decision to disqualify the offender has had access to all this info that you and Egor posted. This info wasn't shown anywhere to someone who is new to the discussion. In this case, it's fine to disqualify.

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9 лет назад, # ^ |

+24

maybe he is multi-threading human? :D

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pedromnasc

9 лет назад, # ^ |

← Rev. 2 →

+132

tourist also should be punished... it's clearly impossible to solve 7 problems in 1:08 alone.

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fake_delete_account

9 лет назад, # ^ |

-6

He can :D

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9 лет назад, # ^ |

-23

He won two ICPC world finals and in the second one his team solved all 13 problems, are you still sure that it's impossible for him?

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pedromnasc

9 лет назад, # ^ |

I was just kidding :)

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9 лет назад, # ^ |

-12

Ok I'm sorry, I didn't notice :)

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EvgeniSergeev

9 лет назад, # ^ |

+21

Wouldn't be the first time he single-handedly beat a whole North Korean class.

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Edvard

9 лет назад, # ^ |

+56

Maybe they don't know that contest for only one coder?)) Or maybe they need a good result to be not killed by their authorities?

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9 лет назад, # ^ |

Maybe :D

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9 лет назад, # |

+37

I love this contest

very nice problems.

Hello div 2 again.

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9 лет назад, # ^ |

same here,especially problem D had nice statement.

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P_Nyagolov

9 лет назад, # |

+60

D was truly written in an AWFUL way, it took me around an hour to get the statement behind the simple problem :/

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9 лет назад, # |

+172

k = n in C is so wrong in so many ways >_>... Who would want to play a game which is already ended? I think such tests should be excluded since we have "Stannis starts the game." and when it's already ended it's false xD.

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vitux

9 лет назад, # ^ |

← Rev. 2 →

+43

I'm also very disappointed about that.

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9 лет назад, # ^ |

exactly! how could Stannis start the game if the game never starts!?

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Reyna

9 лет назад, # ^ |

i just found out i got hacked cause of that and now i got accepted :|

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9 лет назад, # ^ |

well I was in a really quiet room. No hacks in my room!

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jason_yu

9 лет назад, # ^ |

But sadly it was stated that K <= N. That was really tricky. What a sad story!

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9 лет назад, # ^ |

-14

If there would be a constraint that k ≤ nⁿ then it would also be true — constraints given by inequalities in algo problems stay for conditions which are necessary for test to be correct — not sufficient ones.

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tomlau

9 лет назад, # ^ |

Sounds similar to "stressful victory" in this years FBHC, doesn't it? :D

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9 лет назад, # ^ |

Is there a chance they change the test cases and rejudge C solutions!?

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9 лет назад, # ^ |

Frankly saying — of course no. However I will stick to my understanding that k = n should be forbidden, but someone can always use some lame excuses like "statement was vague and since it wasn't excluded explicitly by constraints then you should consider this case" (which is not a valid reasoning of course :P).

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9 лет назад, # ^ |

Can't agree.

Formal > non-formal, so input > description.

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9 лет назад, # ^ |

-13

See my comment above: http://mirror.codeforces.com/blog/entry/18348?#comment-233274 Moreover, how "Stannis starts the game" is non-formal for you?

There is no such thing like "input is more important than description" or other way round. All of them should be consistent and rules assured throughout whole statement are on the same level of importance.

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9 лет назад, # ^ |

Consider this version of actions during the game: on their turn a player checks if there are just k cities left; in such a case, the game ends and otherwise, that player makes a move. For-loops work this way: the ending condition is checked before the first iteration.

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9 лет назад, # ^ |

Actually there's a lot of problems that have test cases where the game already ended before it starts

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9 лет назад, # ^ |

How is that justifying anything :P?

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9 лет назад, # ^ |

You said test cases such that the game is already ended should be forbidden when the problem statement says that "[name] starts first" , but a lot of game problems exist on codeforces or other OJ stated that "[name] starts first" and has test cases such that the game already ended before it start.

so I meant to say that it's very normal thing to have test case k=n in this problem just like it's normal in other game problems

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9 лет назад, # ^ |

It's a reasoning with which someone may disagree, but it is a valid reasoning. The statement may be considered vague (at least I needed to translate from Weird English to English sometimes in this contest) in such a way that the case k = n wasn't explicitly excluded. If a case isn't explicitly excluded by the statement, one should consider it (I didn't). Which part of this would be objectively invalid?

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9 лет назад, # ^ |

"Which part of this would be objectively invalid?" — in an "ideal mathematical model of statements" there is no distinction between "explicitly excluded" and "implicitly excluded" and I consider(ed) that phrase as implicitly excluding, therefore excluding.

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9 лет назад, # ^ |

"I consider(ed) that phrase as implicitly excluding"

That's not objective.

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9 лет назад, # ^ |

It is if we will assume that "starts the game" implies "first move exists". One may assume it, one may not, however there should be only one "true meaning" and what we can do now is to debate whether it is as I said or not.

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9 лет назад, # ^ |

← Rev. 2 →

"One may assume it, one may not"

Yep, that's what "vague statement" means. To you, the statement means one thing, so you disgree with the reasoning you call invalid. To someone else, the statement may mean something else — not falsely — and that person would disagree with that same reasoning.

You've made your point and I see your point, but I don't dismiss everything else. The important thing is: what to do next? There's nothing more to say. What there is to do: improve at reading statements, improve at problems of "UGH CASEWORK" type, lead by example by setting problems clearly, offer to review problems for future rounds (if you don't want to participate...) or find someone else who could do that. Path to perfection...

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Hedayet

9 лет назад, # ^ |

The authors might say, "You know nothing(about Stannis's stubbornness)" ;-)

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9 лет назад, # ^ |

+13

If it said "Stannis plays the first move", you would be right, but it doesn't. "Stannis starts the game", and the game immediately ends, no problem there.

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9 лет назад, # ^ |

Hmmm... Probbaly you're right, however whatever the "true meaning" is, I guess it would be best if I will just leave it as it is, stop complaining and be more careful next time :p.

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9 лет назад, # ^ |

He has to make a move to start the game :|

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kojak_

9 лет назад, # |

+37

With all due respect, let's all take English to the grave and bury it together after this contest. Specially on problem D.

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9 лет назад, # |

+18

Am I they only one who had problems with viewing the code for hacking ? The whole code goes out of the box !

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-Wave-

9 лет назад, # ^ |

Yes, but after closing and reopening, it fixed itself.

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9 лет назад, # ^ |

I did that .. but it continued with me for the last 15 minute .

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bluepichu

9 лет назад, # |

+26

Shoutout to dreamoon_love_AA for hacking my C (with n = k) so that I fixed it and got a lot of points on it :)

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manetsus

9 лет назад, # ^ |

Same to me! Thanks dreamoon_love_AA . :)

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shm0007

9 лет назад, # ^ |

Me not so lucky! hacked in 2:28 with case k==n :(

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9 лет назад, # ^ |

Be happy. At least you almost solved a less solved problem!

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Ali1377

9 лет назад, # |

+10

We hope a good stay in San Francisco for tourist!

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9 лет назад, # |

-13

Hoping to become purple, please update ratings fast.

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9 лет назад, # ^ |

And I became purple for the first time!!

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jlahcejlot

9 лет назад, # |

Интересно это нормально ? Test: #54, время: 0 мс., память: 4 КБ, код возврата: 0, код возврата чекера: 1, вердикт: WRONG_ANSWER Ввод -1 1 1 1 Вывод 0.99999997764 Ответ 1.0000000000 Протокол тестирования wrong answer 1st numbers differ — expected: '1.0000000', found: '1.0000000', error = '0.0000000'

Посылал в дорешку. http://mirror.codeforces.com/contest/549/submission/11475797

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9 лет назад, # ^ |

Так у тебя погрешность на самом деле больше 1е-9. Ошибка уже в 8 знаке после запятой, так что вполне нормально.

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jlahcejlot

9 лет назад, # ^ |

Точно, все верно. Перепутал требуемую для задачи точность. My bad, thx.

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knst

9 лет назад, # ^ |

← Rev. 3 →

[комментарий удалён, повтор]

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Kvark161

9 лет назад, # ^ |

да, это нормально, ваш ответ не достаточно точный, а error, expected и found вывелись с 7ю знаками после точки, на самом деле error = 0.00000002236...

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9 лет назад, # |

← Rev. 2 →

how to solve G? I tried by sorting the people according to the initial cash they hold, updating the final cash in hand, and checking for validity. But this doesn't pass the pretests. What is the correct solution?

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Teshnizi

9 лет назад, # ^ |

How to solve E?

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9 лет назад, # ^ |

i see what you did there ;-)

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http://mirror.codeforces.com/contest/549/submission/11476753

9 лет назад, # ^ |

Got my mistake. We need to sort according to a[i]+i

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ivanilos

9 лет назад, # ^ |

I just passed after contest. For every guy in line give him what would be his value if he is last in line. grab the minimum of this values and update the rest of the array. It is safe to put him at the end of the line as no one else can have value smaller than him by definition of minimum.

When updating everyone who was before the guy with minimum value, say x, they paid him one dollar, though their position become closer to the front and the number of guys in line decreased by one, so we subtract one from all these guys.

For every guy who was in front of x also subtract one as the number of guys in line decreased by one.

Then check if we meet the conditions given of the line.

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pwypeanut

9 лет назад, # ^ |

Since the amount of money a person has is "adjusted" by the change in his position, we can sort the list of people by (original money + original position), then we can place the one with the lowest (original money + original position) in front, then the next one and so on, subtracting off the position from (original money + original position) as you place them. If the resultant order is not sorted, then it is not possible, so output impossible.

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nisige

9 лет назад, # ^ |

"then we can place the one with the lowest (original money + original position) in front" in front you mean the end of the line?

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pwypeanut

9 лет назад, # ^ |

Yup, end of line or "front" of output. Thanks for the clarification.

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shef_2318

9 лет назад, # |

-67

Вот почему нельзя в условии B было написать, что человек всегда себе звонит? Как это можно понять не вчитываясь во "Входные данные"? Из самого условия я понял, что могут быть петли, а не то, что они будут обязательно. Есть ощущение, что я такой далеко не один.

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9 лет назад, # ^ |

И сразу же вопрос вдогонку — без дополнительного условия о единицах она вообще решаема за полином?

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9 лет назад, # ^ |

+22

Нет, доказано, что НП-полная, 2222 3-САТ свел к ней.

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9 лет назад, # ^ |

При этом все стараются как можно больше времени потратить на веселье, поэтому рассылают сообщения всем, без разбора, и даже себе.

Это написано в двух местах в условии, в первом абзаце и во входных данных. В первом абзаце даже специально отмечено болдом.

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shef_2318

9 лет назад, # ^ |

-22

Вот из первого абзаца я и понял, что могут быть петли(и на это обращают внимание, поэтому выделили). А о том, что они есть всегда по-моему можно понять только из входных данных.

Мне кажется, что написав бы на пару слов больше(или полпредложения добавив) никому бы точно не стало хуже, а при этом бы такая ситуация ни у кого бы не могла возникнуть.

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vovuh

9 лет назад, # ^ |

Повторю за Максимом с добавкой.

поэтому рассылают сообщения всем, без разбора, и даже себе.

Гарантируется, что i-й символ i-й строки всегда равен 1.

По-моему, тут всё предельно понятно.

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Kostroma

9 лет назад, # |

С каких пор переполнение int в одной из двух самых сложных задач в контесте перестали отлавливать претестами? Спасибо конечно авторам за бесценный опыт, но по-моему это странно.

→ Ответить

9 лет назад, # ^ |

← Rev. 2 →

Мы думали, что там одна сложная задача :(

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riadwaw

9 лет назад, # ^ |

Казалось бы, даже если она не сложная...

→ Ответить

9 лет назад, # ^ |

+34

Если серьезно, то я считаю, что не взять лонги — такая же бага, как не разобрать случай с несвязным графом, с одной вершиной, взять маленький массив, неправильно посчитать eps, сделать мало итераций, etc. По этому поводу можно долго холиварить: кому-то это нравится, кому-то нет. Я считаю, что этот момент остается за автором, который принял такое решение, зная, что проблемсет не содержит кучу реализаций, и, наверное, чуть проще среднего.

Тем более, что в этой задаче лонги очень даже очевидные.

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Al.Cash

9 лет назад, # |

+45

Dropped far from top 20 because added eps for comparison in H. So stupid :(

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matrix

9 лет назад, # ^ |

← Rev. 2 →

+10

Same here.

EDIT: Actually I wasn't in the top 20 anyway. But I think Maybe in the top 200 :D:D

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Sumeet.Varma

9 лет назад, # ^ |

even i added 'eps' and failed. Sadly i also failed for n=k case in C . bad day..i could have been orange easily :(

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sp765

9 лет назад, # |

Sorry if this is a stupid question (first time participating), but: The contests announcement said the order of problems is going to be random, but couldn't people have guessed the value of problem based on how fast its points are falling off? (there was a table on RHS that listed how many points I can get at that point in time for each problem).

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9 лет назад, # ^ |

Yes; that doesn't falsify the statement that the problems are ordered randomly, even though you can figure out what the "proper" order should be by watching the scores falling.

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sp765

9 лет назад, # ^ |

I don't recall exactly what I saw in the table, but:

How did the points all start out at 3000 if some problems were actually worth only 250, 1000 points, etc.? For instance, if problem A showed 2xxx points in the table early in the contest, wouldn't that have been misleading?

Thanks.

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9 лет назад, # ^ |

That is called dynamic scoring. The score of a problem depends on the number of people that have solved it; naturally, the more people that solved it, the less the point value is. But at the beginning of the contest, the number of people that will eventually solve it is obviously unknown (not happened yet), so the scores are based on the current number of people that passed pretests.

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9 лет назад, # ^ |

These numbers depended on the number of solutions that passed pretests at any point in time. You could just look at the dashboard and guess the order of problems based on the number of solutions listed there.

It's perfectly okay to use that.

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9 лет назад, # ^ |

Of course, but it's only a guess. Also, it might be very wrong because everyone submitted a wrong solution (see problem C).

→ Ответить

ashkan_d13

9 лет назад, # |

← Rev. 2 →

One unsuccessful hacking attempt can send someone from 1600 to 2700

As did to me! :(

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PattS

9 лет назад, # |

← Rev. 2 →

+41

Series of silly bug

C: forgot base case where n = k
F: use cin/cout without sync_with_stdio(false)
H: use eps

from ~80 -> 943 Orz

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manetsus

9 лет назад, # ^ |

don't worry, better luck next time.

→ Ответить

tuna_salad

9 лет назад, # ^ |

← Rev. 4 →

Mine:

G: used cout and printf with sync_with_stdio(false)
H: intended to use max{|A|,|B|,|C|,|D|} but forgot ||'s

Good bye, T-shirt.

→ Ответить

Sumeet.Varma

9 лет назад, # ^ |

+13

It feels nice when you are not alone who failed due to silly mistakes!

C : n = k
H : eps

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9 лет назад, # |

+18

Seems that the writer of C scenario hates the Lansters :P

→ Ответить

SpyCheese

9 лет назад, # ^ |

← Rev. 2 →

What will they do if they burn King's Landing (with the Iron Throne)? It will happen if k=0.

→ Ответить

9 лет назад, # |

← Rev. 2 →

Is $\text{[math]}$ an expected complexity to E? I came up with a solution based on inversion pretty quickly, but when I realized that I need to compute convex hulls on fractions (or on doubles and have hope that this somehow will pass) and that TL seems very risky for such complexity (with hevy operations on fractions/doubles) I felt sad :(.

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Endagorion

9 лет назад, # ^ |

I believe O(nm) is possible via randomized half-plane intersection; not sure if this was the intended solution though.

→ Ответить

9 лет назад, # ^ |

No, tutorials will be published very soon and you will see, sorry for delay:)

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9 лет назад, # |

In problem C, what would be the answer for this case:

8 2

2 2 2 1 1 1 1 1

with explanation please.

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trilis

9 лет назад, # ^ |

+13

Daenerys wins, because she can destroy all cities with population 2. Then there will be two cities with population 1 in the end.

→ Ответить

9 лет назад, # ^ |

well, you have other possible scenarios that makes Stannis the winner.

how I'm I suppose to decide ??

→ Ответить

9 лет назад, # ^ |

← Rev. 2 →

In these type of tasks you should suppose that both players choose the best possible strategy.

In this case, if Daenerys plays optimally, Stannis have no way to win.

Another possible explanation — choose the strategy when the only answer exists. If there is no such strategy — the answer is impossible (if the task allows this kind of answer).

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pi37

9 лет назад, # |

Can anyone explain why my G failed? Shouldn't nlogn sorting pass for n<=2*10^5? http://mirror.codeforces.com/contest/549/submission/11470961

→ Ответить

mkirsche

9 лет назад, # ^ |

You probably need to use PrintWriter for output and maybe even a faster input method such as BufferedReader.

→ Ответить

m.khooryani

9 лет назад, # ^ |

use faster input and output methods and store answer and print it once instead of print part of answer each time in a loop

your code with a little change in input and output: Your text to link here...

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DEGwer

9 лет назад, # |

← Rev. 2 →

E: since the values of coordinate are small, we can decrease the number of points by ignoring the points which is not in the convex hull.

After that, for each point we use geometric inversion at the point, then we can easily calculate the answer by sorting the points by the argument.

The time complexity is O(|coordinate| log |coordinate|)

→ Ответить

9 лет назад, # ^ |

← Rev. 2 →

Wow, that is so smart! Problem with large number of points magically disappears.

However I'm pretty sure that removing points from convex hull leaves $\text{[math]}$ not $\text{[math]}$ of them, where z is a range of coordinates.

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9 лет назад, # ^ |

Your claim is correct.

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DEGwer

9 лет назад, # ^ |

+13

Oh, you are correct. I believed it was O(z^(1/2)) until now...

→ Ответить

9 лет назад, # ^ |

← Rev. 2 →

Btw, I also came up with an inversion, but what do you mean by "we can easily calculate the answer by sorting the points by the argument"? In my understanding we need to check whether there is a line that separates two sets of point which is equivalent to checking whether two convex polygons intersect, which is not that trivial.

→ Ответить

DEGwer

9 лет назад, # ^ |

sorry, I mistook. But we can calculate the answer O(z^2log z), by judging whether two convex hulls intersect.(since two polygons are convex, we can judge it by plane sweep method)

→ Ответить

9 лет назад, # ^ |

← Rev. 4 →

Lolololol, that is so wrong, you can't disregard points inside convex hull ( ͡° ͜ʖ ͡°) ( ͡° ͜ʖ ͡°) ( ͡° ͜ʖ ͡°)!! But that was a good try :D. I wonder whether there are tests detecting this :P.

UPD: However we can throw away points inside one of them — that one which we apply inversion to. But this blows up complexity to $\text{[math]}$ :P. But it should pass either way

UPD2: Duh, what I said it's not true as well -_-. I got lost xd

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Nicolas16

9 лет назад, # |

+93

I was at place 212 in the first "final standings" and I am now at place 202! Could you find two more cheaters :p? These T-shirts are so nice!

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vitux

9 лет назад, # ^ |

+93

I was at place 316 in the first "final standings" and I am now at place 311! Could you find 111 more cheaters :p? These T-shirts are so nice!

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Suhaylee

9 лет назад, # ^ |

← Rev. 2 →

+20

And the chance accepted :)

→ Ответить

Nicolas16

9 лет назад, # ^ |

+12

Awesome :D. It makes my day!

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Nicolas16

9 лет назад, # ^ |

Nooo, I am back to place 204 :-(

→ Ответить

9 лет назад, # |

← Rev. 5 →

Прошу прощения, что означает вердикт "Попытка игнорирована"? Я получил "Претесты пройдены", до конца контеста ничего не отправлял больше по задаче, сейчас обнаружил, что попытка игнорирована без какого бы то ни было пояснения. Как это работает?
UPD: http://mirror.codeforces.com/contest/549/submission/11462849
UPD2: Зачли. Но хоть комментарий бы написали, а то меня уже минусуют, как будто у меня сразу было зачтено. Спасибо :)
UPD3: Увидел коммент :)

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Azret

9 лет назад, # ^ |

← Rev. 2 →

Отлично!

Надо ещё найти одного читера для Nicolas16 и 110 для vitux.

P.S Это, конечно, шутка... если вы не читер. :D

→ Ответить

9 лет назад, # ^ |

← Rev. 2 →

+14

Я рад вашему задору, но если вы отвечаете за свои слова, то будьте добры пояснить, в каком месте кода я считерил. Код имеется по ссылке. (Не хотел отвечать резко, но прямые безосновательные обвинения тоже не слишком приятны)
P. S. Рад, что у вас есть чувство юмора :)

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Azret

9 лет назад, # ^ |

Извините если я вас обидел. Видимо мне стоило полностью дописать коммент перед тем как публиковать, но до вашего ответа я успел добавить правку.

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9 лет назад, # ^ |

Мне кажется, все три наши правки были сделаны в тот момент, когда следующий комментарий уже писался, поэтому не были прочитаны вовремя. Своеобразно получилось, да :)

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9 лет назад, # ^ |

Вообще, если причина в том, что моё решение было признано читерским, и если это сделано автоматом, то было бы неплохо, если будет сразу выдаваться комментарий. Ну то есть, типа "Было списано из статьи ... ресурса ..." или "... из посылки ... участника ...". Это должно сильно упростить решение таких ситуаций. Да и вообще уведомление это неплохо. Я мог закрыть контест и открыть через неделю результаты. А то получается как-то очень грубо и без уважения.

→ Ответить

9 лет назад, # ^ |

+33

Я не могу сказать, как так получилось, что у вашей посылки образовался статус "игнорирована", откатил ваши результаты.

По поводу комментария к игнорированной посылке — идея хорошая, но на текущей стадии развития все читеры игнорируются по большей части вручную, поэтому писать такие комментарии пока не представляется возможным в силу огромного количества случаев читерства. Наверное, ваш сабмит был причислен к их числу по ошибке, приносим свои извинения.

Тем не менее, мы работаем над улучшением данной процедуры.

→ Ответить

9 лет назад, # ^ |

Скорее всего, ручная модерация делается на основе выборки, сделанной какой-то программой, мне слабо верится, что есть иной способ. И модератор должен получать от такой программы какую-то информацию о том, почему посылка попала в выборку. Скорее всего, если в случае игнорирования такая информация будет автоматом прокидываться в уведомление, это позволит сильно сократить нервы. Понятно, что информация будет не полной, но хотя бы номер сопоставленного решения будет сильно полезен :)

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Lo_R_D

9 лет назад, # ^ |

Раньше вручную проверяли, но после VK cup 2015 всё автоматизированно.

→ Ответить

9 лет назад, # ^ |

На данный момент все три мои посылки проигнорированы. Узнал я об этом случайно. Зато, похоже, теперь знаю причину. Мне прислали несколько номеров посылок, код в которых совпадает (частично или полностью) с моим. Причём не все они заигнорированы. Думаю, причина в том, что я писал в ideone.com (так уж сложились обстоятельства), а кто-то догадался посмотреть на предмет наличия решений на этом ресурсе в последних сниппетах. Вот незадача. В общем, полагаю, вина в моей безалаберности. В принципе, установить, что код именно мой, не так сложно, просто посмотрев стиль моих предыдущих посылок, но сейчас уже глупо спорить об этом. Так или иначе, но код утёк. Если посылки останутся проигнорированы, то я просил бы сделать контест нерейтинговым для меня -- задачи я решил честно, а получать минус в рейтинг странно из-за этого. Достаточно потраченных времени и сил.

По поводу вежливости хотел бы дополнить ещё раз -- всё было бы куда мягче, если бы изначально было уведомление, а не безмолвное и беспощадное блокирование посылки.

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9 лет назад, # ^ |

Вообще, я пригляделся к человеку, посылка которого совпала с моей. Надо сказать, выдающаяся личность: http://mirror.codeforces.com/submissions/jk95/contest/549

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rivudas

9 лет назад, # |

+62

Now lets move on to the Champions League Final!! :)

→ Ответить

9 лет назад, # ^ |

Which team wins: Barselona or Juventus? :)

→ Ответить

supervillain

9 лет назад, # ^ |

+18

surely juve ;)

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gskhirtladze

9 лет назад, # ^ |

← Rev. 2 →

+16

Barcelona* will win ;)

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pratham2508

9 лет назад, # ^ |

-11

barca most probably

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P_Nyagolov

9 лет назад, # ^ |

← Rev. 2 →

+13

My favourite team is Real Madrid and my second favourite is Roma... It is hard for me to choose which team to support this night and most likely I will just enjoy the game :)

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xlwinner11

9 лет назад, # |

← Rev. 3 →

Может быть я что-то пропустил в условии, но в задаче H нигде не было указано, что элементы вырожденной матрицы, которую мы будем искать, не могут быть нулями. В таком случае, мы можем составить матрицу в которой один из элементов — максимальное значение матрицы, данной во входных данных, а все остальные — нули. Определитель данной матрицы будет равен нулю. Таким образом, ответ всегда 0.

~~Подскажите пожалуйста, где я неправ.~~

Глупость.

UPD: Спасибо большое всем прокомментировавшим: ошибку понял, в следующий раз буду внимательнее.

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J4T8Z9

9 лет назад, # ^ |

Ну, как бы, смотрите.
Нам нужно минимизировать max(a[1][1]-b[1][1],a[1][2]-b[1][2],a[2][1]-b[2][1],a[2][2]-b[2][2]).
Как это, по-вашему, всегда будет равно 0?
По простой логике, если max(a[1][1]-b[1][1],a[1][2]-b[1][2],a[2][1]-b[2][1],a[2][2]-b[2][2])=0, то a=b, по-другому никак. А входная матрица, конечно, не обязательно вырожденная. Перечитайте условие.
З.Ы. 11468397 Хорошая попытка, но нет=)

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xlwinner11

9 лет назад, # ^ |

Спасибо большое, теперь я понял то, что надо учиться верно читать условия.

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DryukAlex

9 лет назад, # ^ |

Так ведь ответ тогда будет не ноль, а максимум из модулей трех оставшихся элементов...

→ Ответить

9 лет назад, # ^ |

Так можно, но это не будет верным решением.

Например исходные данные: (1 1 1 1) Ваше решение: B = (1 0 0 0), k = 1 (наибольшее из абсолютных значений разности исходной матрицы и B) Оптимальное решение: (1 1 1 1), k = 0

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Stochastic

9 лет назад, # ^ |

← Rev. 2 →

Ответом будет норма матрицы A-B, в вашем случае — второе максимальное значение по модулю матрицы А.

upd: Долго писал :)

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9 лет назад, # |

+17

I hope updating the ratings doesn't delay for too long

→ Ответить

DanAlex

9 лет назад, # ^ |

+19

"The final results will be announced in a day, after catching all cheaters. " ... So the ratings will change only after the final final standings ?

→ Ответить

Tamtomus

9 лет назад, # |

First I got accepted on my solution for problem A , then the verdict changed to "skipped" , why ?? Can someone explain for me ? And I've only submitted ones.

→ Ответить

Darkmaster

9 лет назад, # ^ |

If you share the submission link perhaps someone can help you :)

→ Ответить

LiTi

9 лет назад, # ^ |

-11

Perhaps you shared your code with someone else...

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Fcdkbear

9 лет назад, # |

+81

Очень классные думательные задачи, спасибо авторам!

→ Ответить

huge_kuriboh

9 лет назад, # |

-12

Wish you'd try to catch cheaters every round because many people participate together. Why don't you add a report option? If someone spotted 2 or more codes very much the same they could report them to you.

→ Ответить

mruxim

9 лет назад, # |

+45

I wish after 49, there was 50.. Where happiness meets cottons :)

→ Ответить

9 лет назад, # ^ |

-9

That definitely sucks

→ Ответить

Teshnizi

9 лет назад, # ^ |

← Rev. 2 →

+12

Top 4 contestants(sorted in unluckiness order!)

1.scott_wu

2.zemen

3.You!

4.kostroma.

→ Ответить

9 лет назад, # ^ |

+13

If I'm not mistaken, scott_wu works in Sillicon Valley, which is right next to San Francisco, so I wouldn't call that very unlucky xD

→ Ответить

aliasadiiii

9 лет назад, # ^ |

← Rev. 3 →

+11

I think he is big loser of the match. :D

UPD : now you become big winner of the match. :))

→ Ответить

LiTi

9 лет назад, # ^ |

+23

they've got 2 cheater below you, congrats ;-)

→ Ответить

LiTi

9 лет назад, # ^ |

Sadly, it seems that they were not cheaters and you're 51 again :'(

→ Ответить

anzu

9 лет назад, # |

← Rev. 6 →

-28

deleted

→ Ответить

Animal.Toad

9 лет назад, # |

-8

Will be there any editorial, plz?_

→ Ответить

Darkmaster

9 лет назад, # ^ |

← Rev. 2 →

-12

Here it is: Editorial

Edit: ¿Why the downvotes?

→ Ответить

9 лет назад, # |

-11

I have a question. In the final standings, it doesn't show that I solved A. It just shows I have 1 unsuccessful attempt at D. Why?

→ Ответить

9 лет назад, # ^ |

Now it shows your Accepted A.

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9 лет назад, # ^ |

No it still doesn't. http://mirror.codeforces.com/contest/549/standings/page/48

→ Ответить

Dedushka

9 лет назад, # |

+29

Посмотрите, пожалуйста, что у меня в решение не так? Не проходит третий тест(( http://mirror.codeforces.com/contest/549/submission/11474914

→ Ответить

9 лет назад, # ^ |

Судя по логике Вашего решения, Вы полагаете, что игрок, который ходит последним, проигрывает, если нельзя уничтожить города какого-то типа. На самом деле, все наоборот. Да и странные Вы какие-то условия проверяете — противоречивые и непонятно откуда полученные. Советую прочитать разбор.

→ Ответить

huansuz1

9 лет назад, # ^ |

+114

Дедушка, кажется это не ваше решение.

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ashkan_d13

9 лет назад, # |

← Rev. 2 →

-17

→ Ответить

9 лет назад, # |

+24

Dynamic scoring in contests with joint divisions is very poor. C worth 2000pts is a joke. As a result we get people with F done and without n=k case in C ending up lower than those with case n=k considered.

→ Ответить

9 лет назад, # ^ |

-9

Because of the number of solving users, I have not read the problem statement E and F. And I spend a lot of time for problem C and calculate as more as possible on my A4 papers. However, when I see others solutions for C when hacking, I was shock for the length.

→ Ответить

9 лет назад, # |

+19

Thanks for the contest. This contest shows my poor English reading skill. I spent nearly an hour in struggling with the statement of D. Thanks for the example input and output, and strict pretest of D.

→ Ответить

9 лет назад, # ^ |

← Rev. 2 →

BTW, who can give me some tips or advice to improve my English reading ability in Programming Contest?

→ Ответить

Darkmaster

9 лет назад, # ^ |

As in every other language, the way to improve is practicing. If you want to improve your understanding of the English statements then the best way is reading and trying to understand lots of other problems, out of competition, here on Codeforces

→ Ответить

9 лет назад, # ^ |

you can refer to the answers on this question on quora: link

→ Ответить

http://mirror.codeforces.com/contest/549/ratings/page/1?order=BY_RATING_CHANGE_DESC

9 лет назад, # ^ |

thx

→ Ответить

ashish1610

9 лет назад, # |

← Rev. 3 →

Sorry its already mentioned in the post.

→ Ответить

dkirienko

9 лет назад, # |

+21

Очень интересно выглядит таблица изменения рейтинга по итогам контеста.

shad0w_walker имел до начала контеста рейтинг 1518, набрал на контесте 0 баллов, и получил +261 к рейтингу (не решив ни одной задачи!). И таких пользователей, получивших большой прирост к рейтингу имея 0 баллов, немало.

Но вот следующий за ним Ivy_End имел до начала контеста меньший рейтинг 1399, решил больше (1890 баллов), а прирост к рейтингу у него всего лишь +258. Возможно, тут дело в том, что у Ivy_End до этого был всего лишь один контест, но можно найти и пользователь, у которых достаточно много контестов написано, первоначальный рейтинг был меньше, чем у shad0w_walker, решили они существенно больше нуля задач, но не получили такой сильный рост рейтинга.

→ Ответить

dkirienko

9 лет назад, # ^ |

А, это читеры с проигнорированными посылками:

http://mirror.codeforces.com/submissions/shad0w_walker/contest/549

Сначала им пересчитали рейтинг, а потом уже заигнорили на читерстве. Может быть тогда и изменение рейтинга откатить?

→ Ответить

MrDindows

9 лет назад, # ^ |

Рейтинг пересчитают после определения всех читеров и объявления окончательных результатов

→ Ответить

dkirienko

9 лет назад, # ^ |

Посмотрел таких, у которых стоит 0 баллов и большое изменение рейтинга.

Из тех, у кого указана страна в профиле подавляющее большинство (десятки) из Китая и Индии. Пару раз попались египтяне, один раз — Кипр, один раз — Казахстан, один раз — Белорусия.

→ Ответить

Erilyth

9 лет назад, # |

Why were some people's codes skipped? Although the codes were skipped their ratings still increased.

→ Ответить

SomeRandomGuy

9 лет назад, # ^ |

Cheaters

→ Ответить

9 лет назад, # ^ |

If you have submitted two or more solutions and get pretest passed, only the last time will be system test, and the others will be skipped.

→ Ответить

9 лет назад, # ^ |

← Rev. 2 →

i think Erilyth is referring to something like this. UPD: i think the issue has been fixed.

→ Ответить

webstorms

9 лет назад, # |

Can anyone explain in detail the binary search solution to question H? I have been struggling for the last hours to solve it/understand it.

→ Ответить

starius

9 лет назад, # ^ |

My solution: 11477556

Binary search for the answer (maximum for absolute value of change of any of a, b, c, d). Minimum possible value is 0.0. Maximum possible value is max(min(|a|, |d|), min(|c|, |b|)). (Because it is sufficient to replace one of a, d with 0.0 and one of c, b with 0.0.)

Step of binary search. Find maximum and minimum for a * d and for b * c, changing any of them (a, b, c, d) up to x, where x is current value of the answer. See my functions bestMax and bestMin. They select the best of 4 values. (Proof: the result of a multiplication is a liner function of each of the values multiplied. Maximum value of a linear function is observed when its argument is min or max.)

Then you compare four values. If range of (a * d) values intersects with range of (b * c) values (see function best), then such values can be selected that (a * d) — (b * c) = 0. If this condition is satisfied, then the answer is <= current value of x, set max bound of binary search to current x. Otherwise set min bound of binary search to current x.

Number of steps is set to 1000, so precision is about a value of 2^1000, which is much more than 10^9.

→ Ответить

webstorms

9 лет назад, # ^ |

Thanks a lot, could you tell me what is wrong with my code? http://mirror.codeforces.com/contest/549/submission/11489337

→ Ответить

starius

9 лет назад, # ^ |

Your code works with two minor fixes: 11492104

replace "Double.MIN_VALUE" with "l1 * r1". Double.MIN_VALUE is the smallest positive value. Results of the multiplication can be negative, so Double.MIN_VALUE is not suitable.
replace "lower < upper" with "lower <= upper". The value for which lower == upper must be accepted. Otherwise the error may be greater than 10^-8 (see my attempt 11492063 )

→ Ответить

webstorms

9 лет назад, # ^ |

Thank you! What is the significance of l1 * r1?

→ Ответить

starius

9 лет назад, # ^ |

"l1 * r1" is one of values. Other option is to use Double.MAX_VALUE * -1

→ Ответить

sazerterus25

9 лет назад, # |

Wow, just wow!! I must applaud codeforces on the algorithm they are using to catch cheaters. But the algorithm is so effective, that it is not only catching catchers, but also innocent problem solvers. Kudos to codeforces!! -_-

→ Ответить

9 лет назад, # |

+25

Когда примерно появятся окончательные результаты и "правильный" рейтинг?

→ Ответить

MikeMirzayanov

9 лет назад, # |

+141

We've finished to process similar solutions. This time we unrated users with very-very similar solutions, but didn't count the first (by time)/original solutions in each equivalence group. Next round we will punish them too. We believe that it is the duty of a participant to not allow solutions to leak. For example, you should not use ideone with enable Public - your code will be available to everyone feature. Also next time we will make deeper analyze to compare solutions and find cheaters

In case of repeating cheating an account will not be able to take part in rounds for some time.

Please work personally. Use of another's solutions does not make you smarter. It is ugly and shameful, it is disrespectful to Codeforces and other participants.

→ Ответить

P___

9 лет назад, # ^ |

-24

Could you Please explain how the current dynamic scoring works?

→ Ответить

rajat1603

9 лет назад, # ^ |

+56

Problem A was a cakewalk problem and it had several (about 3000+ ) solutions accepted . So chances of 2 or more solutions being same are really high , i personally know some people who solved it really quick and i am sure they didn't cheat but still they got caught for cheating and this round went unrated for them , i think an exception should be made for problem 'A' only and should not catch cheaters for this this problem . Thank you :)

→ Ответить

animeshf

9 лет назад, # ^ |

+31

I am in a similar situation. I submitted within ~9 minutes and I think it is the reason I have been disqualified from the contest :/

→ Ответить

sazerterus25

9 лет назад, # ^ |

-8

It looks like my soltions got accepted again. Sorry for the angry outburst, but I do think you should look into the solutions more carefully.

→ Ответить

boleyn.su

9 лет назад, # ^ |

← Rev. 2 →

There are also cheaters who have two accounts and use one to view other's solution (after passing the pretest) and then use the other one to copy it. To handle this situation, I think problem writers should make sure their pretest are strong enough.

Another solution is to add more limit on hacking. For example, one can lock a problem if and only if his solution have passed the pretest for 30 mins or the contest will end in 30 mins.

→ Ответить

9 лет назад, # |

cheaters removed and Update again rating and hello div 1 again (:

→ Ответить

9 лет назад, # |

+112

Однако мой код доступен всем в моей комнате, кто решил эту задачу. И если кто-то решит дать списать своему товарищу, он может просто заблокировать задачу и дать ему чужой код. В таком случае будет дисквалифицирован невинный человек?

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Julek

9 лет назад, # |

← Rev. 3 →

+238

Losery Cup 2015

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9 лет назад, # ^ |

Roman Giertych, is that you?

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9 лет назад, # ^ |

← Rev. 2 →

+15

Good news. Maybe another cheater is found. You are the 200-th now! Congratulations!

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NikaCh

9 лет назад, # |

+16

I didn't get green t-shirt but I got green :)

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A.Elahi

9 лет назад, # |

+75

I'd rather have my username exactly the same on the back of my t-shirt. I mean Monyura shouldn't be monyura on his t-shirt.

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9 лет назад, # ^ |

Sure. Monyura is Monyura on CF only. In all other places he is monyura. So, it's just a bug on T-shirt example.

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9 лет назад, # ^ |

+33

Don't worry, we will use Case-sensitive designer)

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ZJiaQ

9 лет назад, # |

I had taken part in the contest but I can't find the changing about my rating.I solved 5 problems,and one of them fst. but now I can't find any record about it.who can tell me why

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9 лет назад, # |

А что с призами? Не видел ни одного комментария организаторов по этому поводу. Они будут разосланы по почте, на адрес, указанный в профиле? Свяжутся ли со мной, когда посылка будет отправлена?

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SpyCheese

9 лет назад, # ^ |

Хороший вопрос. Что с призами?

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9 лет назад, # ^ |

+20

Вам, как и всем призерам, должно было прийти письмо от MikeMirzayanov с просьбой проверить, что поле адресс находися в актуальном состоянии. Если это так, мы напишем сообщение, как только приз будет отправлен. Сейчас мы ведем переговоры, как лучше эту отравку орагнизовать, все призы найдут своих героев:)

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dragoon

9 лет назад, # |

+51

Did any one get their tshirt?

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zemen

9 лет назад, # ^ |

+24

Same question. Neither t-shirt, nor prize.

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9 лет назад, # ^ |

+21

So, is there anybody who got their prizes now?

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Al.Cash

9 лет назад, # ^ |

I did :D

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9 лет назад, # ^ |

+31

I didn't get my prize yet T.T Is it only me who couldn't receive the prizes?

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I_love_tigersugar

9 лет назад, # ^ |

+13

Me either :(

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9 лет назад, # ^ |

+10

Finally I got my T-shirt today!.. but not the Ollie thing :p They received my address about a month ago, so I think I should wait for another 9 months to get the Ollie :D?

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jcvb

9 лет назад, # ^ |

I also got my T-shirt yesterday :)

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I_love_tigersugar

9 лет назад, # ^ |

I also received my T-shirt today. So nice :) Hope further T-shirts will have strange colors like this :)

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8 лет назад, # ^ |

+25

6 months passed after I got an email about the prize, but I have no idea where the prize is. Is there anyone else who still couldn't get their prizes?

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8 лет назад, # ^ |

← Rev. 2 →

I still haven't received an ODK. I sent multiple reminder emails, and they stopped responding to those. Can someone help us here?

EDIT: This has been resolved and I received an ODK.

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