Всем доброго времени суток!
22 сентября 2015 года в 19:30 MSK состоится очередной раунд для участников из второго дивизиона. Традиционно, участники из первого дивизиона могут участвовать в соревновании вне конкурса.
Авторами раунда являются я (Владислав Вишневский) и igdor99(Игорь Дорошев). Хотелось бы сказать большое спасибо Zlobober(Максиму Ахмедову) за помощь в подготовке задач, Delinur(Марии Беловой) за перевод условий на английский, MikeMirzayanov(Михаилу Мирзаянову) за замечательные системы Codeforces и Polygon, а также нашим друзьям daksenik(Дмитрию Аксенику) и irevt(Ивану Ревту) за помощь в подготовке раунда. Это наш первый раунд, и надеемся, что не последний!
Вам будет предложено 5 задач и 2 часа на их решение.
Главным героем раунда является попугай Кефа, любящий деньги и рестораны.
Всем удачи и высокого рейтинга!
UPD: Разбалловка — 750-1250-1500-2000-2500.
UPD: Разбор!
Почему раунд перенесли с 20 на 22 число? Надо было что-то переделать?
он по времени совпадал с раундом codechef
Ну чё ты начинаешь, нормально же общались(((
Забавно
Видимо эта рассылка совсем автоматическая, то есть она берет имя автора последнего поста на главной =)
3..2..1..Go!
Summary: Vladik and igdor99 would like to please Zlobober, Delinur, Mike and RevtIvan. It looks like they will have their hands full for a while.
I'm sure they will all be pleased if the round goes well :)
I assume Jefe tried to go for an obvious innuendo. I learn more about CF community every day.
In part I wanted to point out they made an English mistake, swapping "Thank" for "Please".
Why no div1 :/
And in the hope ... That i shall get some hacks :D GL & HF :)
And in the hope ... that i shall get some hacks :D GL & HF :)
Oh it will be a magical round.
P.S : Please see the tags.
And it's also a magical number of the round (321).
thanks for magical contest about parrot KEFA)
Do you mean Kefa ?
i just want raising rating .......
Unfortunately, the probability of all users get raising rating is zero :(
hope to see hard math problems :)
Maths problems is sometimes interesting(brainstorm). But too many maths problems will make my brain feel tired.
Занимаешся футболом и любишь матиматические задачки????
Страный ты однака...
i agree
i too
Div. 1 guys do not make new accounts please. :)
Toooo_Simple
Why did you make a new account then ?
You're unrated and you know that some of div1 contestants make new accounts and you know the whole story, Then definitely you're not new to Codeforces and I can say that you already have another account!
Why did you?
The one who is doing that is a retarded person!
I hope that it does't be dynamic. How about you? Do you agree with me?
we want div.1 contests .
My last contest in summer! I'm going to school on Wednesday! Hope a good contest! Good Luck!
but from your profile it looks like this will be your first contest.
this will be My first contest but My last Summer contest , ok ?
Oh My God :) , are You sure this contest your first contest? But I think this contest is your first contest by fake Account
Раунд 321: цифры по убыванию
капитан очевидность
Почему перенесли раунд с 20.09 на 22.09?
попробуй прочитать первые комменты
Все, понял, спасибо!
I am no longer scared of fake contestants. If my rating goes down, its because I suck, and not because of anyone else.
Yes the sucker man!!!
Lol
Get better than their main accounts and take their real rating as revenge!
Hello my friends ;)
Check the 1st version of his comment . He called everyone b*****s >_<
It was just a typo my friend ;)
Ugh!!! Don't you just hate it when you write "friends" and codeforces autocorrect changes it to "b*****s" ? :p
dqehbfqeadnq;ekhfadxnqekfnasxdmajkefnjadlnwe
Scoring will be 500-1000-1500-2000-2500?
no
In my whole country adsl is down. I'll be solving this contest on mobile net :)
Good luck all CF guys!!!
750 points for div2 A :| from where that it cant be very nice i think it will be a long boring brute force :(
This can also mean difficulty of A,B,C is not necessarily in that order.
Why doesn't codeforces allow a little bigger hack files? There are solutions that will get time limit exceeded when the constraints are used but the files are bigger than 256 KB.
You must write a test generator
There's only field for writing the test case, right?
In such case, you should write test-case generator.
MLE on pretest 1 in D :(
What for the sake of God is wrong with pretest 9 Div 2 C?
Isn't it finding the longest non zero subsequence from an arbitrary vertex
Iwith no childs?i believe there was situation when parent has larger index than a child
Damn, should have thought about this. I shall try after the system testing.
I don't know maybe my approach is wrong but I run DFS to see if we can reach at leaf without consecutive vertices's containing cat or not and add it to the result! But my code was failing on test case 8 and 9
My code was failing on test case 8 This code http://paste.ubuntu.com/12523672/ and after swapping the index of the x and y when x > y My code failed on test case 9 http://paste.ubuntu.com/12523669/
just add an extra line:
to:
and I don't know what the
if statmentwhile reading really does!!My wrong assumption — If 1 is leaf, don't count it in answer. I kept counting it and kept getting failed submissions.
How can 1 be leaf when 1 is root and n>2
1-2 2-3 3-4 3-5
1 is root and a leaf(in a way)
How is 1 a lead in a way? I has 2 for a child
Draw it on a paper, ad see for yourself. Clearly, assuming 1 to be leaf was wrong n this problem, but not in general, when the tree is unrooted. That's why it was failing.
Leaf is a node with no child. 1 has a child 2. Seems like you got bit confused about leaf.
I suppose yes. I used the condition 'only 1 adjacent node' to identify leaf.
How can 1 be leaf when 1 is root and n>=2
A lot of pretests, yet smooth system testing and no issues. Really nice contest.
How to solve E problem?
It was so interesting.
I passed pretests with hashes + segment tree for updating hash value in a range.
Segment + Hash
I pass the pretests by compressing all values to bits.
I thought about this and hashing also, but I couldn't figure out a way to check the query correctness except by iterating on each block and check if the hashes/bits equal.
How to do this part quickly?
String s[l, r] has period d if s[l, r-d] = s[l+d, r]. You can get the hash of a substring with a segment tree.
I feel so stupid right now ._.
That's brilliant, I've never seen this trick before.
Except when d = r — l + 1. Nice observation though :)
I get TLE after system test. XD
In my room, MrDindows successfully solve it by treating all character as long long. You can refer his code: http://codeforces.net/contest/580/submission/13165100
Hey, can you explain his code. He is using memset and that is O(n) if I am not wrong. His solution seems to be O(n) per query.
You can find many information about memset from Internet.
Lets ignore the memset part. His query is still O(n).
He do some thing to speed up the constant part of his algorithm such as treat character as long long and expand the while loop.
I guess these changes of code will make CPU execute lesser command. (I'm not familiar with the detail, but I know these are common trick to speed up your code.)
How to deal with cyclic dependcy in D ?
Since the constraints are small, it can be solved by dp + bitmasks.
Does anyone have any ideas how to solve Div2E without using hashing?
Ignore... It's wrong.
Why lcm(nextk1, $next_{k2}
Unable to parse markup [type=CF_TEX]
) | d ?What do your ki means?
Sorry, I just figured out it's wrong.
how to solve E ?
Опять пол раунда лагал кф. Открыть чужое решение было нереально в течении некоторого времени.
P.S. как взламывать решения генерируемым тестом, в частности что значит "Параметры командной строки"?
P.P.S. Только через минуты 2 после конца контеста, мне показало, что мой взлом прошел.(хотя, я его пытался отправить минут 5)
Pretest 9th for C ?
Maybe here
I was unable to submit in the last 60 seconds because codeforces was too busy. I hate it when that happens.
And I was unable to hack in the last 60 seconds.
Same here :/ I tried to hack in the last 20 seconds, but the challenge wasn't submitted.
same as your problem :|
solved D but It didn't submited, also in previous contests I have this problem too!!!
same here...
Блин решил последние две задачи а кодфорцез затупил и я не сумел послать(((((((((( уверен, это дело рук красных они бояться конкурентов и поэтому зажимают F5 на последних минутах и кодфорса ломаеца. предлагаю ограничивать красным доступ на сайт на время контестов.
I don't like problem scores!!! u think difference between problem C (or B) and D should be ONLY 500 (or 750).than it's same to solve A+B instead of D. May be my fault coz I solved only A,C,D(coz of not enough time),but anyway I don't think that this correct: man who solved first 3 problem is in front of me in list... D is much harder than A+B+C -_- sorry for my open mind
What's the problem with code generated tests ? I've got this one:
Thanks in advance.
validator wants '\n' in the end of a test.
Check this out.
Your test has extra space
Error: space is in the end of the second line.
It's correct:
Thanks, guys. That's just hilarious ;-(.
Спасибо за контест, мне он понравился
Is there anything need to pay attention to in the C problem? I failed to pass the 8th test and i can't find any mistakes.
Me too. :(
I just see the test data and I found it was because I wrongly think xi and yi in input mean xi is yi's father node.
You assumed given graph(tree) to be directed.
Can someone suggest an approach for 2nd? I basically sorted the friends based on their money, and I was then going to group friends based on the condition and sum their friendship factor, and check which group has the most. Is there a better way to do it? or any flaw in my solution.
you can read about the two pointer technique , and try solving the problem.
Firstly,sort the friends based on their money.Then,enumerate every element in array and use the function "lower_bound" to find the first element which is bigger than the it.And use the sum of factor between the two element to update your answer
1) sort based on money
2) find prefix sums for friendship factors
3) for i=1 to n
a) use any O(log n) mechanism to find index j such that money[j]>=money[i]+d ... I used c++ map for this.
b) ans = max (ans,prefix_sum[j-1]-prefix_sum[i-1])
I had the same solution. Used
std::lower_bound(a + i, a + n, a[i] + d)to find the least index ahead ofiwhich I can not select, if I selecti.Index j is also invalid so why are you doing prefix_sum[j-1] and not prefix_sum[j] ?
Am assuming that i > j as we are considering the ith element and the old set if in [0..i-1]
Assume your set starts from i. We have sorted in ascending order of money so j>i because j is the lowest index you need to exclude if the set starts from i such that money[j]>=money[i] + d ... you can set money [n+1] = INF ... Include all elements from i to j-1 in current set and to get the sum of friendship factors of such ranges/sets of in O(1), I pre-calculated prefix sums. To get sum of elements from i to j-1, you need to subtract prefix[i-1] from prefix[j-1].
Sort + Two Pointers O(nlogn)
Can someone explain D soln ?
I solved it by DP, take a d[n][2^n] array,where d[i][j] means what is the best answer when last dish is i and j represents which dish I tasted before it,(1 means tasted,0 not tasted)
dp[mask][j] — best answer if we finish eating at j — dish, and ate all dishes in mask. O(n^2 * 2^n)
constrains of n & m are small. I think it can be solved using recursion trying all possible ways.
UPD: it will pass only by using memoization..
That would be about N!-(N-M)!, which is too much.
Aren't all possible cases O( P(n,m) ) ? (Which should be too high for recursion)
I'm sorry for this. I updated it.
and here's my solution: http://codeforces.net/contest/580/submission/13170384
Well, I think it is almost the same as the Traveling Salesman Problem
You have a graph, where the dishes are the nodes and the eating rules are the edges
dp[y][bit_mask]:
y is the last node you visited (dish you ate)
bit_mask is bit mask in which if the Nth bit is set it means that somewhere along the path you visited node N (ate dish N)
so
dp[y][bit_mask] = max (dp[x][bit_mask — (1 << y)] + eating_rule[x][y] + a[y])
if there isn't a eating rule between x and y than eating_rule[x][y] should be zero
so your final answer is the best dp[y][bit_mask] where bit_mask contains exactly M ones
you have N * 2 ^ N states and the you need O(N) for each, so total complexity is O (N^2 * 2^N)
Used same idea, got AC in Practice. Link to Code
Went out and could not fully participate in contest. Too bad!!!.
My thinking of how to solve Div2 B.(not sure if its correct)
Suppose we have <n=k people already to be a company. They are K-1 of them in number.
Consider the Kth person.
Case 1 : He should be in company. This is only possible if this amount is a < S[0].a That is it is smaller than the amount of the smallest person in the group.
Case 2: His amount is bigger.
In this case, we can use binary search to find a range of numbers that conflict with the Kth person, we sum the total factors of this said group.
We have to choose between the whole group and the Kth person. Which ever increases the overall amount.
Please would this have solved the problem???
I got AC using this approach
OMG!!!, I can't believe i just threw away my chance of turning blue today. Oh well, i'm happy i got the approach at least. Lets wait for the next round.
Thanks for the reply Expert Katalonecfly.
В чём смысл давать задачи, в которых не надо думать вообще? Задачи A-B должны быть легкими, с этим никто не спорит. Но даже легкие задачи можно попытаться сделать интересными.
Но ничего не меняется и дальше:
C — напишите DFS.
D — напишите дп по маскам.
E — чуть-чуть подумать надо, но всё же понятно, что единственный способ решить задачу с такими запросами — это дерево отрезков (или какое-нибудь другое) над хешами.
Контест кончился.
Да, задача E сложная, но думать в ней все равно не надо. Достаточно сложно придумать идею ДО над хешами, думаю, что людей, способных на такое, единицы. Так что либо ты знаешь эту идею и для тебя задача выглядит как "напишите такую-то структуру данных", либо не знаешь и не сдаешь.
Справедливости ради — вот Е без хешей, 13165100 :)
Ну это вообще проблема див2 раундов, они в 90% случаев страдают от того, что слишком стандартные. Т.е. div1C/div2E в div1 раундах преимущественно более идейные, чем div2 E в div2-only.
Разве
memsetне даст TL при1e5запросов на забивание всей строки циферками?А почему это заходит?)
1010 операций в 1 секунду, очевидно же должно заходить =)
Еще чуть-чуть поизвращавшись (можно паковать по 2 символа в 1 байт), получаем AC за 327 мс со все той же квадратной асимптотикой.
Копелович отдыхает :)
Когда читаешь комментарий и решение и понимаешь, что следующий Всерос точно не сольешь
A — напиши стандартную штуку, или минус задача, если не умеешь
B — напиши стандартную штуку, или минус задача, если не умеешь
C — напиши стандартную штуку, или минус задача, если не умеешь
D — напиши стандартную штуку, или минус задача, если не умеешь
E — напиши стандартную штуку, или минус задача, если не умеешь
Сегодня рассказывал коллегам, как валить хеши по большому модулю. Придумал, как делать для двух разных модулей и оснований, написал прогу, которая проверила гипотезу. Работает! Эх, думаю, хорошо было бы, если бы на вечернем КФ была какая-нибудь простая задачка на хеши, я бы вдоволь поломал.
В общем, спасибо авторам за задачу E. И nfssdq за ресабмит с другими основаниями и возможность взломать его ещё раз, что стоило мне первого места.
А ведь когда-то солнце было ярче, трава — зеленее, а хэши даже в случае модуля 2 в 64 никто не ломал.
А теперь пишешь хэши по простому модулю, а их все равно ломают — получается, нужно теперь привыкать к тому, чтобы дописывать туда рэндом? :) Или как от такого защищаться?
Я честно сделал оба основания рандомными, зависящим от входной строки. Собственно, вот: 13158976.
Вроде бы я умею объяснять, почему по рандомному основанию сломать нельзя: степень многочлена — длина строки, то есть корней не больше, значит, оснований, по которым будет ноль, не более n, и вероятность угадать мала, а в случае двух разных оснований ещё меньше.
Впрочем, осталось научиться реверс-инженирить сишный рандом, и можно становиться богом взлома хешей.
http://codeforces.net/blog/entry/4900 ?
Нет, я методом Капуна делал.
Что значит "для двух разных модулей и оснований" ? То есть ты умеешь ломать хеш по любому двойному модулю?
Умею. Выше Um_nik кинул ссылку, как можно делать, а я метод Капуна адаптировал.
А достаточно менять только основания и не трогать модули? Я почему-то всегда делал наоборот (основания разные рандомные порядка 10^9, Q = 239017), но сейчас задумался, что это должно быть не оптимально.
Can't wait to see the rating change!!!
good picture
were the test case for 2nd question weak because i used upper_bound instead of lower_bound in hurry and i got AC? :P
I used upper_bound in one failed submission. You're lucky :)
I know why it passed coz i upper_bound on make_pair(curr+d,0). If instead of 0 it would have been a greater value it would not have passed. Yes I was lucky, coz i did not see that through.
PS : the test cases weren't weak then i guess.
The test cases were not weak, but solutions that did not consider test cases similar to yours passed. Overall, the solution is still correct and adding these test cases would've made the problem unnecessarily tricky.
I have this idea for D. Please tell me if you find something wrong in it.
Is it correct to start from leaf nodes?
I don't know if it's correct but it's an overkill.
Ok ;_; . I'll look at the editorial then.
It is dp with masks. State is last eaten and bitmask of eaten so far. To solve a state you try to fix the next to be eaten.
Oh my god, i have disastrous bugs :'( i will cry in the bathroom.
I assumed that input was (parent, child) on C!!! ... silly mistake.
I agree with everything above and confirm that I will provide my own problems
Just a quote regarding D.
How can we solve problem B if every friend was associated with hia own value of 'd', the value that makes him feel poor.
Sort + fast search algorithm.
Here!
So if you fix a[i] as being the biggest of sums,you need to add all a[j]<a[i] that a[j]+d[j]>a[i].
This can be done with fenwick tree,at position a[j]+d[j] you add factor[j].
And for i the answer is query per range [a[i]+1,limit]
Or linear parsing, with two pointers
http://codeforces.net/contest/580/submission/13150523
Nice problem ( E — Kefa and Watch ). I found a solution using hashing + segment tree, but couldn't solve it in time. BTW is there any solution without using hashing?
kuldeepfouzdar nice trick for Unsuccessful hacking attempts on your solution, in the problem B:
#define int long longhttp://codeforces.net/contest/580/submission/13163755
Хотелось бы извиниться за столь баянные задачи, но все таки не стоит забывать, что эти задачи были подготовлены для второго дивизиона. Если я не ошибаюсь, этот раунд стал рекордным по числу зарегистрировавшихся пользователей, и я действительно сожалею, что не смог оправдать чьих-то ожиданий.
P.S. Если вам это поможет, поставьте этому комментарию
-.Hello all,
I have a question about an unsuccessful hack that I did during the contest. I hacked the following submission as I noticed that the numbers of the sequence are stored in indices of an array starting from 1 to n. The array is of size n and hence I expected that the hack will result in run time error due to the index overflow for a large input where the size of the sequence is 10^5.
Can anybody explain to me why did the hack fail? I expect it might have something to do with the unused array f but I am not sure about it.
C++ doesn't have index overflow run time errors, all that happens is that it accesses memory outside the array and as long as said memory is not used for something else it will pass.
How come 300 div2 participants managed to solve D? In my opinion it was a pretty hard dynamic programming problem.
Really ? I think that's the basic form of bitmask dynamic programming though
Basic bitmask would not have arbitrary transition rules, most in div 2 fails at problems like this. I think its strange that it was solved by as many div 2 participants as this problem which is much easier:
http://codeforces.net/contest/574/problem/D
This problem is also easier and was solved by just 100 div 2 participants:
http://codeforces.net/problemset/problem/540/D
So I'm just wondering: what makes this problem easy?
Rather than "easier", I would say it is more well known
IMO, one of the most basic problem for bitmask DP is finding shortest path to travel through all vertices in the graph which is completely the same as 574D, except that the answer lie only in status
2^n - 1P/s : this is the first bitmask DP problem I learned so it can be just me
UPD: I've seen some tutorials about bitmask DP (just google search, you'll find them) use above problem as sample to solve so it should be well known
Ah, I see! Thanks! I had just seen it for things like edit distance and knapsack and those are way easier.
Future div1 participiant ;)
Good job, there is no way I could have solved that when I was in div 2!
Stand-up contest)
Really nice contest, can't wait for your next one.
When will the solutions be uploaded?
There are solutions in the editorial=)
so many O(m(n+k)) solutions passed E 13198157
In Problem B, I think it should be "Print the maximum total friendship factor that can be reached." in the Output statement.