Codeforces Round #226 (Div. 2)

11 лет назад, # |

Мне вот интересно, если див 1 участвует вне конкурса, могут ли они делать взломы?

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riadwaw

11 лет назад, # ^ |

Да, у них отдельные комнаты.

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11 лет назад, # ^ |

круто, спасибо

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NVAL

11 лет назад, # ^ |

Да, внутри своей комнаты с такими же участниками из div1

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worse

11 лет назад, # |

-20

I want to rating under color of my eyes... (red)

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Ali1377

11 лет назад, # ^ |

+13

You say this in all contests?! :D

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Eran

11 лет назад, # ^ |

+16

i think i've heard this earlier

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Ali1377

11 лет назад, # ^ |

Take a look at this.

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caustique

11 лет назад, # |

+38

I was always curious why not just to add 2 more problems and make Div1+Div2 round instead of just Div2 one. It seems to me that the level of Div1 problems is not always very high — for example, take a look at D and E from previous round 225. They're just medium, not hard and still were not solved by a lot of contestants. I mean, one doesn't need to overestimate level of Div1 participants but it would me much cooler to have more frequent Div1 rounds.

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Killever

11 лет назад, # ^ |

Big Like

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alohamaloha

11 лет назад, # ^ |

Just like topcoder.

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11 лет назад, # |

-17

why +5 min ?

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knst

11 лет назад, # ^ |

← Rev. 2 →

+26

All right, I was late, but now I'm here. Start! (just a joke) :)

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nikalosaberidze

11 лет назад, # ^ |

Don't worry Lasha !!!

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BiMokh

11 лет назад, # |

-12

5 minutes delay?
Why!?!

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mory91

11 лет назад, # ^ |

+19

maybe he forgot to write the last problem! :D

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ans_

11 лет назад, # |

Страсти накаляются

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SC_pRo_ION

11 лет назад, # |

А! О! Какой ужас! Перенесли на ПЯТЬ МИНУТ! Уписаться.

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11 лет назад, # |

Как решать C ? и что там за четвертый тест?

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htkek

11 лет назад, # ^ |

В четвертом тесте, некоторые L и R > 10^7. Нужно учесть этот вариант.

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snake13

11 лет назад, # |

Как делать C без факторизации всех элементов массива? 1731 мс как бы намекают...

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11 лет назад, # ^ |

-9

По всей видимости, никак.

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noogler

11 лет назад, # ^ |

-8

можно. Если предпочитать для всех чисел наименьшее простое число в разложении.

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11 лет назад, # ^ |

Ну подсчитали (по всей видимости, линейным решетом) и что дальше? Вероятно, как раз таки факторизовать эти числа.

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noogler

11 лет назад, # ^ |

← Rev. 3 →

leastprime[i]= наименьшее простое число, вот таким образом делаем факторизацую

while (x > 1) {
   primes.push_back(leastprime[x])
   x /= leastprime[x].
}

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11 лет назад, # ^ |

Это и есть факторизация.

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noogler

11 лет назад, # ^ |

извиняюсь я кажется подумал что с факторизации :(

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11 лет назад, # ^ |

Фрагмент кода вашей программы:

	while (x > 1) {
		int p = prime[lP[x]];
		if (ls != p) cnt[p]++;
		x /= p;
		ls = p; 
	}

Что это, если не факторизация?

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11 лет назад, # ^ |

у меня такая же идея, по почему то не пашет 4 тест((

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caustique

11 лет назад, # ^ |

Just calculate for each prime number of its multiples that are in the given set. The complexity will be N / 2 + N / 3 + N / 5 + N / 7 + ... = O(NlogN) where N is a maximum possible number (10⁷).

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caustique

11 лет назад, # ^ |

+24

Я не знаю, что на меня нашло, но я ответил по-английски, хотя знаю русский:)

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enot110

11 лет назад, # ^ |

+35

На самом деле, там чуть лучше асимптотика ;)

$\text{[math]}$

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11 лет назад, # ^ |

← Rev. 2 →

вот http://codeforces.net/contest/385/submission/5787970

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ELKIN_

11 лет назад, # |

Solution for B with n^2 complexity passed my hacking test :'(

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11 лет назад, # ^ |

← Rev. 2 →

+13

B is supposed to be solved in $\text{[math]}$ . ofcourse it will pass the hack!!

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logic_max

11 лет назад, # ^ |

+23

There is an O(N) solution also.

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11 лет назад, # ^ |

can u plz share the idea with us?

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logic_max

11 лет назад, # ^ |

For each occurrence of "bear", you have to count the number of valid indices to the left of this, and to the right. It will contribute countLeft*countRight number of tuples. For each "bear", left count will be the number of characters from 'b' of previous "bear" to this 'b'.Right count will be number of characters from this 'r' to the end of string.

You can check my submission(5788550) for details.

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11 лет назад, # ^ |

← Rev. 3 →

thanks a lot! :)

EDIT: i wonder how your $\text{[math]}$ submission and my $\text{[math]}$ submission (5791720) both pass with a time of 78 ms :D

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alv-r-

11 лет назад, # ^ |

I got TLE at first with O(n²), then got AC with O(n) :(

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Metamorphus

11 лет назад, # ^ |

← Rev. 2 →

Man, just a piece of advice for you. On Codeforces if you wanna hack somebody on time limit, you gotta be sure, that his/her program will perform at least 10^9 operations on your test. Because the servers are really fast. On this contest I tried to hack a program, which performed in the worst case about 4*10^8 operations, but the result was just 1668 ms.

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omse1998

11 лет назад, # |

Из-за лагов на последних секундах не смог перепослать задачу :/

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BiMokh

11 лет назад, # |

Couldnt submit my code in last 20 seconds :-@
Again loss of rating :-(

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jibon_ebong_shomoy

11 лет назад, # |

at last minute codeforces was unavailable, So I couldn't submit my soln.Did someonne else have this problem?

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MohammadFatafta

11 лет назад, # ^ |

Yeah same thing happened here!!!

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adelnobel

11 лет назад, # |

Couldn't submit a problem in the last two minutes because of constantly getting the message "Codeforces is temporary unavailable "

I really hate this!!!

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jibon_ebong_shomoy

11 лет назад, # ^ |

Me too -_- .. spent 1.5 hours for 1 problem , couldn't submit it.. what a joke ..

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kojak_

11 лет назад, # ^ |

I complained about this. Answer: "No comments"

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kuruma

11 лет назад, # |

Hello,

I need guidance on problem B.

Can it be solved by suffix array?

I thought of using suffix array + lcp to get all substrings only containing bear and count those.. Maybe even suffix tree was better choice but couldnt implement it yet.. was that the way to go?

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ELKIN_

11 лет назад, # ^ |

nope, solution is very simple. For each "bear" count positions on left from which we can start, it will be (i — prev 'bear' position), because we shouldnt use the positions already used by previous bear and positions on right on which we can end, it will be = n — i — 3. then increment ans to rightcount*leftcount, and change last bear index.

sorry for bad explanation. if you disunderstand then look to others code

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Mostafa_36a2

11 лет назад, # ^ |

← Rev. 5 →

I've BruteForced it , and it passed

Edit : TLE in testcase 9

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AU.Bahosain

11 лет назад, # |

Codeforces is temporary unavailable :(:( It cost me hack in the last minute :(

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princeladdak

11 лет назад, # |

Whats the idea of problem C ,i got TLE.

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11 лет назад, # ^ |

you need to use sieve of Eratosthenes in liner time

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Mediocrity

11 лет назад, # ^ |

+14

I got AC with simple sieve of Eratosthenes.

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http://codeforces.net/contest/385/hacks/91276/test

11 лет назад, # |

+27

I liked problems very much,exceptionally problem C,thanks CleRIC for great contest ;)

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PraveenDhinwa

11 лет назад, # |

I wasted 10 minutes to write a generator. It only gives the error that first line should not be empty.

Can anybody help me why is this happening ??

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vintage_Vlad_Makeev

11 лет назад, # |

+34

Спасибо Вам! Отличный раунд!

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PraveenDhinwa

11 лет назад, # |

For problem D constraint n <= 20 suggested that something like bitmask dp could be done. But I can not find any reason why not to use just all the flash lights instead of a subset of them ?

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kingofnumbers

11 лет назад, # ^ |

Not sure but I think this is because the order of lights matters.

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PraveenDhinwa

11 лет назад, # ^ |

← Rev. 2 →

I got the idea, infact it is true that all the lights can be used. But the bitmask can be used to extend intermediate result dp (mask) to dp (newmask) easily using geometric manipulations.

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Uni

11 лет назад, # |

+36

My hack #91170 was marked as unsuccessful. But the details contains "Time: 2152" and this problem (Problem C) has 2 seconds time limit. Doesn't 2152 > 2000?

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11 лет назад, # ^ |

← Rev. 2 →

in the last two contests, i saw problems with time limit of 2 seconds, and submissions that took 2011 ms got accepted.

wait 2-3 mins, i will post the links.

EDIT: sorry for the delay, but finally found them! 5722902 5706257

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MikeMirzayanov

11 лет назад, # ^ |

+12

I'll handle it. Already has found the issue. Thanks.

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numa

11 лет назад, # ^ |

← Rev. 2 →

There may be something wrong.
I found a successful submission for problem C which takes 2245ms to execute. 5796553
And I hacked a code which causes TLE, according to verdict, it ran 2000ms.

To keep contests fair, this problem should be deeply discussed.

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Ali1377

11 лет назад, # |

What a "runtime error" contest!

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11 лет назад, # |

How solve problem E?

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Andrei1998

11 лет назад, # |

What was the idea behind D? (n<=20 sugests that the problem is NP for bigger values of n). Was is something like meet in the middle?

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oml1111

11 лет назад, # ^ |

More like travelling salesman dp. For every subset of lamps store the farthest point that can be reached using them. When adding a lamp to a subset it is easy to see, how much it extends the reach.

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pbtbbo597

11 лет назад, # ^ |

I just solved D, the dp[] size is (1<<n), than every step only turn on one unused light and update dp[cur|(1<<k)] (k is the unused light)

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tinybells

11 лет назад, # |

For the first time, my rating is going to go above 1500 (hopefully) :)

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karthikabinav

11 лет назад, # |

Any idea why http://codeforces.net/contest/385/submission/5798194 gives a run time error. But uncommenting the commented lines doesnt give. I cant see what fails.

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p-space

11 лет назад, # ^ |

← Rev. 4 →

yes, i have the same mistake, the thingis that s.size() is an unsigned int... and when it is 1 and substract 3, it is 4294967294, because all the expresion is indeed unsigned int, and then when you access to the element 4294967294 of the string it give you RE, because it only has 1 element.

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karthikabinav

11 лет назад, # ^ |

ohh interesting. yeah this makes sense. thanks !

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mohamednabil00000

11 лет назад, # ^ |

I think there is nothing wrong on your code. but the compiler make infinte loop but I don't know why.

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ghooo

11 лет назад, # ^ |

← Rev. 3 →

because s.length() is unsigned int

and by negating 3 from it, it will give another unsigned int, so

s.length()-3

will be some non-negative number

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Kianoosh

11 лет назад, # |

← Rev. 13 →

+97

Rank 2:SquirrelDetected Registered: 4 hours ago(Before Contest)
Rank 3:MahooshojoNHB Registered: 7 weeks ago(Before Contest)
Rank 4:Dong5k Registered: 4 hours ago(Before Contest)
Rank 6:Uni Registered: 4 hours ago(Before Contest)
Rank 7:FastYes Registered: 3 hours ago(Before Contest)
Rank 9:akatsukiLH Registered: 9 hours ago(Before Contest)
The NEW brains...!

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KiAsaGibona

11 лет назад, # |

Can Someone Explain problem E in details (i.e : How the matrix was built?? )

// So far I go...
   |  1 0 1 0 1 0 |     | sx |
   |  0 1 0 1 1 0 |     | sy |
   |  1 1 1 0 1 0 |  *  | dx | 
   |  1 1 0 1 1 0 |     | dy |
   |  0 0 0 0 1 1 |     | M  |   M= Number of moves so far
   |  0 0 0 0 0 1 |     | 1  |   1= Constant term to increase M

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adelnobel

11 лет назад, # ^ |

Well, instead of the first 4 zeros in the last column, they all should be 2. I got a WA during the contest because of this stupid bug. Note that you need to decrease sX, sY by one to turn the coordinates to 0-index instead of 1-indexed to make it easier, consequently, you will need to add 2 to K in each step.

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KiAsaGibona

11 лет назад, # ^ |

Why they first 4 zeros need to be 2 ? What is the logic behind this?? I don't understand.. And when sX or sY is greeter than N we need to mod them by N ... how to apply mod N operation in matrix multiplication?

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adelnobel

11 лет назад, # ^ |

← Rev. 2 →

Well, we will be working with 0-indexed coordinates, ok? Now, dx is updated as follows dx = dx + sx + sy

But since we are working with 0-indexed coordinates then this should be dx = dx + sx + sy Why? Assume sx == 1 && sy == 3 in 1-indexed mode, now when we turn it to 0-indexed then we have sx == 0 && sy == 2, so dx should equal sx + sy + 2

Now for modding by N part, you can mod every thing by N in every step of your calculations, because the only operations you do are addition and multiplication, and the mod can be distributed over these operations normally

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11 лет назад, # |

-10

MikeMirzayanov, пожалуйста, добавте возможность скачивать тесты или каким-то другим образом просматривать весь тест, ибо уже не первый раз по началу теста невозможно определить как он выглядит, почему решение валится и что делать((

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enot110

11 лет назад, # ^ |

+25

писать стресс!!!)

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