Всем привет!
В это воскресенье, 21-го марта 2021 года в 13:00 по московскому времени начнется Финал Технокубка 2021!
Для тех, кто хочет посоревноваться на тех же задачах, будет проведено два обычных раунда Codeforces: один для первого, другой для второго дивизиона. Раунды начнутся 21.03.2021 16:20 (Московское время), не пропустите!
Конечно, если вы участвуете в финальном раунде Технокубка, то вы не можете участвовать в раунде вечером. Мы просим участников официального финала воздержаться от обсуждения задач в открытых сообществах до конца раунда вечером.
Задачи для вас готовили: Александр Golovanov399 Голованов, Евгений amethyst0 Белых, Андрей AndreySergunin Сергунин, Алексей Aleks5d Упирвицкий, Diego Diegogrc Garcia и я.
Также спасибо Bench0310, kokokostya, Um_nik, dorijanlendvaj, brunomont, Stepavly, antontrygubO_o, JinhaiChen, budalnik, wucstdio, golikovnik, kuviman, dantrag, BledDest, Supermagzzz, ruanxiaoyu, geranazavr555, divanik, psevdoinsaf, Roms за тестирование задач и ценные замечания, а также antontrygubO_o за помощь в проведении зеркальных раундов.
Удачи!
Поздравляем победителей в раундах на Codeforces!
Div. 1:
Div. 2:
Всем спасибо за участие! Разбор в обычном формате, презентация с разбором.
You missed " Notice the unusual timing "
You missed your brain
Ok, thanks for letting me know
Друзья, а что такое "технокубок"? Как я понимаю, это олимпиада. Ещё есть возможность регистрации?
Нет, финальному раунду предшествовали отборочные. Подробнее тут.
Hoping for strong sample tests this time ;)
dude ur timeline is chaotic , can u tell desired rank to get to 1400+
1400
Depends on how many participants participating in the round generally you need to be in the top 20% maybe and it differs when it's div 1+2 just being in top 30% would get you to specialist easily
Is the official Technocup round (not the Div 1/2) rated? (not a bait question, genuinely curious)
Should be, all previous finals was rated
The official contest is in a group rather than on the main CF contests page which is why I was wondering...
That answered my question I guess, goodbye Master.
3 contests a day. Gonna be rough. - Kickstart - CF - Cookff
An ARC as well
same like like kickstart round A 2020 (kickstart -> atcoder -> codeforces) isnt it?? and worst of all that time too it was sunday :|
leetcode as well.
CodeChef too
AtCoder Regular Contest 115 — 5 PM (UTC + 6)
CF DIV 1/2 — 7.10 PM (UTC + 6)
MARCH COOK-OFF 2021 — 10 PM(UTC + 6)
Last year, the contest was ethical and beautiful. Hope it will be the same this year!
Today full of contests will coming...
JOISC 2021 Day2 1p.m. — 6p.m.
AtCoder Regular Contest 115 7.p.m. — 9p.m.
CF Round #709 Div. 2 9.10p.m. — 11.10p.m. (Isn't it used to be 9.05 p.m.? Wondering for it)
I've never participated in so many rounds in one day. It's busy without doubt, and I hope it'll also be a fulfilling and unforgettable day for me...
P.S. UTC+8
You forgot about Google Kick-start!!
This round is rated or not...as it's not mentioned anywhere in the announcement ??
Yes this round is rated
At least try to change it a bit when you copy a comment. UPD: Now it's better.
Codeforces Rounds (those that start with Codeforces Round #???) are always rated.
Do converse also true XD?
.
Shouldn't it be Technocup 2021?
P.S. If I misunderstand this, can someone please explain it to me?
I guess that they forgot to update the almost-copy-pasted part XD
KAN please look into this.
Узнают ли финалисты разбалловку до начала раунда?
жаль что некоторые финалисты будут сдавать с фейков:(
KAN а почему финалисты могут зарегаться на DIV2?
Мы их оттуда уберем
Atcder round will end exactly 10 minutes before the start of this round. Such a productive day
And this round ended 10 minutes before the start of Codechef Cook-Off :o
Why unusual time though, so as to host it with the actual finals?
What is the score distribution for the tasks?
How many problems will there be?
Its a surprise.
How Many Problems will be there and problem ratings ???
Problem *Scores
delayed 10 minutes
Delayed by 10 minutes
Forget to release score distribution and make a 10 mins delay?
Come on everyone and good luck!
It's my first time to have a contest!
there is a delay!
SO now what should i do for next 10 mins of my life?
plz dont simply copy paste someones else comments. this comment was written by me in the last edu round.
You can sit and pray the round starts after this 10 minutes.
It's a good idea to stay hydrated!
See how many of your friends are online and taking part in this round? :P
Or remind your friends that you are online.
Delay it some more times and the unusual timing will become usual again.
what's the scoring distribution of the problems?
I guess score distribution is left for surprise.
So, the revealing of score distribution is gonna be the latest in codeforces history, lol!!
Please don't delay this round more than this, we have to give cc march cookoff as well :( T_T
You know they don't delay it for fun, right?
Yeah mate, I know . My apology for being a bit rude :(
by mistake i locked my wrong solution instead of write due to technical glitch during start of contest will i get points of problem A as there is correct one but locked one is wrong please respond whosoever knows????
Once you locked your solution for A , the next solution of A will not be judged by the judge for competition,instead the judge will just take the response which u submitted just before locking. :-|
Why score distribution for this round is not given
Literally guys , again 10 minutes what to do for 10 minutes i controlled washroom for the contest Now fucked up !!
Man just go to the washroom, it's not worth it.
Every minute seems like an hour now :)
exactly man!
Score distribution?
I guess it will be a surprise
no notice for unusual timing no score distribution whats going on?
[deleted]
How can someone solve A,B,C within 2 minutes unless they know the problems and the solutions beforehand! Ridiculous!!!
Its clear they already know the questions. See this guy what a legendary speed he has https://mirror.codeforces.com/profile/OrleanMagic
wtf?
In two minutes this guy read 3 problems as well as solved them and then coded them.
They are geniuses bro
The No.2 Solve A and C in less than 1 minute ...?
That's so strange, at least for me I can't type that fast...
because its based on technocup, which has already ended at the moment of beginning of the d2 round and they might be its participants. i guess they will get punished somehow.
They have been removed.
how about dr_haider ? he also solved 2 problems in 2 mins
[Deleted]
Total solve for each problem not visible on the dashboard(Div2)
Same for Div1
504 Gateway Time-out
Unrated again?
Now I face the problem The statement is not available now...
Down only at the beginning. Later it was fine until the end of contest (at least for me).
No, m1/m2/m3 were working fine for the whole round.
But it lacks some useful functionality. E.g. statistics of solved tasks (one can determine which task better to solve next) and hacking (but it was too rare in this round).
Statistics of solved tasks was available in the bottom of the standings page
Idea to make
m[X]for all users mandatory during the beginning of the round, e.g. first 30 minutes or so, when most participants upload their A and platform overloads.I am submitting using m3 and not able to see if my solution is submitted or not.
It is being unethical and ugly. Again.
Hey guys I can't see the statements of the problems now. What's wrong?
Its not visible in m1 but visible in codeforces.com
Thanks very much! Now I got the statements
Toughest Prob B ever!!
B was pretty hard if not familiar with modular arithmetic. C was easier than B for me.
MikeMirzayanov There were a few people who read , coded and submitted about 2-3 problmes within a first few minuites, I feel thy already knew the problems. Please do us a favour and disqualify them.
They have been removed.
I submitted A within 7-8 min but the page wasn't loading so I resubmitted it on m1.codeforces. The same code without any change was submitted. Both the submissions passed but the later solution timing was considered and I was given some penalty for two submissions with exact same code. Check it and if possible resolve the rating changes.
bro....I lost around 200 points due to this....
Pathetic Contest for me IMO.. No offense to anyone
This is the actual power of your speed
Rank 1591 with A solved
Also Rank 9160 with A solved
:D
Yeah before I solved B I was once 3300 while others who only solve A too got rank 500...
in problem b when i 1st time submited it gives me RE..i tried to find error but couldnt find it..so i tried to submit the same code added some comment line(because you cant submit same code)..and i got WA for same code..What happening :")?
Submit again , you might get AC :D!
if you have written on c++, it's probably undefined behavior
by mistake i locked my wrong solution instead of write due to technical glitch during start of contest will i get points of problem A as there is correct one but locked one is wrong please respond whosoever knows????
Yes, if it passes the system test.
hardforces!! ( atleast for me don't know about others )
Yes!!
For $$$Div.1$$$ it was more like implementation-forces
Really interesting problems, have no idea how to solve them though lol
Really nice problems, no idea how to solve them though lol
I want to clarify 1 one thing,i made submission of problem A at 7 min from m1.codeforces ,and after submitting B from codeforces.com,i saw A has not been marked green ,while its pretests are passed,again i submitted from codeforces.com,then WTF,my score was reduced from 486 to 218,seriously dude,is this fair??
And there is one more strange thing-
I submitted A and B both in m1.codeforces.com and I found that A didn't mark green(Passed pretests) but B did. Now I'm worried that if I won't get any scores in A :(
See standings page to confirm. That page was working fine.
Yeah but the only problem is that I can't lock my problem :(
Mine is also not marked green. It must be a technical glitch, but you should not submit it again. This problem may be happening with many others.
So you can't resubmit it during the game, right?
You can submit as many times as you want. But you should not, as it leads to increased penalty also.
And ya, the same code can not be submitted multiple times.
Most recent submission time is taken into account.If pretests are passed and you submitted again then it counts as "Some edge cases striked you ,which now u think your submitted code wouldnt pass system test and thats why u resubmitted" and thats why most recent time is taken into account,it seems.
same :(
Due to technical difficulties, some problems may be shown as unsolved. However, on the tab "My submissions" you still was able to see your submissions.
but I wasn't able to lock the problem for hacking due to that :D
I submitted A at 14 minutes 50 second and italso passed pretests, but it doesn't show green colour and option to lock in Dashboard. Though, it is there for other 3 problems.
Problem D: fix a source vertex $$$u$$$ and let $$$M = \max_v l$$$ over all triples $$$(u, v, l)$$$. Add an $$$(n+1)$$$-th dummy vertex to the graph, and for each triple $$$(u, v, l)$$$ add an edge from $$$v$$$ to the dummy vertex with weight $$$M - l$$$. Now an edge is useful if it is in a path of length at most $$$M$$$ from $$$u$$$ to the dummy vertex. Now the problem reduces to doing two Dijkstra's (one from $$$u$$$, one from the dummy vertex) and checking each edge. Since the graph is complete in the worst case, it was best to implement the $$$O(n^2)$$$ version of Dijkstra (heap-based version gave TLE for me). Final runtime $$$O(n^3)$$$.
I used a somewhat different approach with the same complexity and was hitting some MLE problems because of heap instead of TLE.
Why not Floyd?
Are you asking if there is an alternative solution using Floyd's, or saying there is one?
There is one.
See my code for details.
KAN What the hell is pretest 16 in D?
How to do div2D?
can you explain div2 B,C?
Each elements is only removed at most once. Therefore, we keep two
std::sets, one to keep track of the remaining elements, and one that keeps candidates for deletion. When deleting a vertex, we can do so by anupper_boundoperation on the second set, update the next vertex that is not deleted.Complexity: $$$O(n \log n + n \log max(a))$$$.
Code: 110638935
Is it rated ?
for sure
I literally wrote some random shit in C and it passed pretests. Waiting for fst :(
How to solve div2 B and C?
Problem B
If there is a solution, then, for all i > 1:
If ai-1 <= ai then ai - ai - 1 = c
If ai-1 > ai then ai - ai - 1 = c - m.
Let c1 = ai - ai - 1
If the values of c or c1 are inconsistent for different values of i then there is no solution.
If either is unknown (i.e. the array is monotonically increasing or decreasing) then any value of m works.
Otherwise the only possible value of m is c - c1. If this is greater than the biggest value in the array it is the solution; if it isn't then there is no solution.
How did you arrive at : If $$$a_{i-1}$$$ $$$ \gt $$$ $$$a_i$$$ then $$$a_i - a_{i-1}$$$ = $$$c - m$$$ $$$?$$$
ai = (ai-1 + c) mod m
ai = ai-1 + c - k*m for some integer k >= 0, by definition of mod
Also 0 <= ai < m
We know 0 <= ai-1 < m and 0 <= c < m, so
0 <= ai-1 + c < 2*m
So 0 <= k*m < 2*m
So k = 0 or k = 1
If k = 0 then ai = ai-1 + c
If k = 1 then ai = ai-1 + c - m
Thanks, I was writing $$$a_i = k*m + a_{i-1} + c$$$.
Takeaway: $$$ a_i = (a_{i-1} + c)$$$ $$$mod$$$ $$$m$$$ is not the same as $$$a_i \equiv (a_{i-1} + c)$$$ $$$mod$$$ $$$m$$$
Any hints for Div2 D ?
Linked list & TreeSet
queue
Duh, I got down to 107 queries in E >_> (worst result after simulating tens of thousands of random tests locally). Is 105 provably optimal (in which case I will admit I was not close) or did I lacked just a few small tweaks :|?
I sincerely hope it wasn't, because otherwise I spent 1.5 hours for nothing :(((((( My approach was to find largest $$$t$$$ s.t. worst case asking $$$(l+r) \cdot t$$$ and always getting lower answer fits in current number of remaining questions. What was yours?
That's what I did as well. I passed pretests (who knows about systests...) by using this approach, running it on an interactor I wrote, and tweaking it in order to maximize the minimum number of queries left on $$$2^{46} - 1$$$, $$$2^{45},$$$ or $$$2^{46}.$$$
On the beginning I tried doubling my budget till I hit first bust ("Fraudster" response). At that point my search space is of form $$$[L, R]$$$, where $$$2L \ge R$$$ (which means that after failed query I can regain a big part of my budget by asking L). Then I try to do skewed binary search. If I ask about midpoint of my interval then no matter what the answer is, I am left with interval of the same length, however in one of these cases I lost a lot of money, so this is probably not the optimal dividing point. I ask about $$$\frac{L \cdot \phi + R}{1 + \phi}$$$ (where $$$\phi = 1.618...$$$). I either end up with a longer interval after gaining money or shorter interval after losing money. Don't ask me why $$$\phi$$$, I just somehow felt it would be good and my experiments confirmed it was a sensible choice.
I was thinking that something like that might pass, but I was assuming that my apporach is a generalization of your's, therefore I would just write mine. I guess it turns out that my approach suffers greatly because of precision, therefore it didn't work
upd: turns out $$$(l+r) \cdot t$$$ isn't equivalent to $$$\frac{(l \cdot t + r)}{1 + t}$$$.................
It seems that after raising number of testcases to a few millions my code got up to even 110 queries. However I added an opt suggested to me by kabuszki that if I have a lot of money I can ask closer to half, if I have not a lot, I ask closer to L (I threw in some random constants to quantify this) and after that it doesn't exceed 104 queries on a few millions of cases.
Yes, I passed with that idea.
How to solve Div 2 E?
We can do this with DP. Let $$$dp_i$$$ denote the maximum beauty you can get by taking photos of only the first $$$i$$$ buildings and $$$bef_i$$$ indicate the greatest index $$$j \lt i$$$ such that $$$h_j$$$ < $$$h_i$$$ or $$$-1$$$ if no such $$$j$$$ exists. Now the transitions are quite simple.
$$$dp_i$$$ = $$$max(\displaystyle\max_{bef_i \leq j \lt i}dp_j + b_i, dp_{bef_i})$$$
The first part is when building $$$i$$$ is in a different photo than building $$${bef_i}$$$, and since all buildings in the photo which the building $$$i$$$ is present are taller than it, we will only consider the beauty of this building. The second part corresponds to the case when building $$$i$$$ is in the same photo as $$$bef_i$$$. In this scenario, as all the building after $$$bef_i$$$ are taller than it, none of the buildings there will contribute to the beauty of the photo. You can implement this using segment tree. Also, make sure to handle the case where $$$bef_i = -1$$$
Can you explain what you will be using the segment tree for here?
For the range max queries in the first part of the $$$dp_i$$$. If implemented naively it might end up with a $$$O(n^2)$$$ complexity. You can use a segment tree or you can use a stack as mentioned in the below comment.
I think it should be max( max ... )
Yep, thanks. Updated.
Thanks for your solution.
I implemented it without segment tree: maintain a monotonic stack. After each operation, the last item in stack stores $$$(i, \displaystyle\max_{bef[i] \leq j \lt i}dp_j)$$$, the second to last item stores $$$(bef[i], \displaystyle\max_{bef[bef[i]] \leq j \lt bef[i]}dp_j )$$$, and so on. It can reduce complexity to $$$O(n)$$$.
Today at starting site was again too slow and when i submitted A it does't reflect in submission so I opened lightweight version of CF and submitted there but later i came to know previous was also submitted ans 50 points got reducted for resubmission :/
I did the same thing and gained -50 points. My bad :(
Can anybody give me some hints for B.
This one passed pretests. Nothing much to explain.
Hint: If an array is valid, then every subsequent difference d=a[i]-a[i-1] will be c if d is positive, and m-c if d is negative.
What about if the array has only negative delta, ie all steps are decreasing?
then m can be largest possible
Example, a[]={ 5, 4, 2 }
Which c, m would make this work?
won't work. 4 — 5 != 2 — 4
Ok, what about { 5, 3, 1}?
(6, 4), (7, 5), (8, 6) and so on
In this case the answer is 0.
Suppose array is decreasing, then a[i — 1] — a[i] = di. And di is same for is i(otherwise answer will be -1), di > 0. Now d = di(any). also, d = m — c. Here m can be any value, c can have any value because d > 0. (so c can have value of [0, m — 1]). So we can any pair of m, c. So finally answer will be 0.
It's still true. The next observation is that if either all are negative, all are positive, or all are 0, then there are no restrictions on m.
I couldn't solve the problem when array has only negative delta For example if the array is 10,8,6,2 like this.
In that case, if adjacent differences are equal the answer is 0, otherwise -1.
if d is negative, then why can't
2m-c,3m-cor in general,Zm-calso one of the possibilities?In our case, the difference between every consecutive number will have magnitude at most m (since each number is between 0 and m-1). c<m so every possibility you listed has magnitude more than m.
The lightweight websites m1, m2, m3 were a bit too lightweight this time...
Same happened with me, but when I reloaded CF official page, it worked.
Such hard problems.
This contest was brutal(Donno about others but for me it was).. T_T :(
I always like these type of contests which is based on some contests. They usually contain nice problems.
Obviously problems were nice, but this type of nice problemset is inversely proportional to +delta. (atleast for me) T_T.
in div 2 e :
for 4th sample why answer is 96?
if we make photo in this way:
1)1,2,3,4,5
2)6
3)7
4)8,9
5)10
beauty is 100
am i wrong?
yes beauty is -8 + (-16) + 4 + 12 + 3 and it's not equal to 100
1) 1,2,3,4,5 -> -8
2) 6 -> 4
3) 7 -> -10
4) 8,9 -> 12
5) 10 -> 3
total beauty is 1
tnx
Why Div. 2 A was so easy?
it was a trap
Does submitting two solutions incur a penalty? I don't know how maybe due to server issues my solution got submitted twice to problem A within 9 seconds and got a 50 point penalty. Any idea why?
Yes there is a penalty of 50 points for resubmission. You can message the contest setter telling him/her that it was by mistake you submitted the same code. He/She can help you.
If the code is exactly same, codeforces doesnt allow you to submit.
I will link to my submissions when codeforces removes the restrictions. Thats why I said it is most likely due to a server side bug. Here are the submissions : Submission1, Submission2
Same thing happened with me
My solution for D:
Let $$$w_{uv}$$$ be the weight of the edge between vertex $$$u$$$ and $$$v$$$ (or $$$\infty$$$ if it does not exist), $$$d_{uv}$$$ be the shortest distance of $$$u$$$ and $$$v$$$ (or $$$\infty$$$ similarly), and $$$l_{uv}$$$ be the third element of one of given triplets that forms $$$(u, v, *)$$$ or $$$(v, u, *)$$$ (or $$$0$$$ if such triple does not exist).
For each $$$(u, v)$$$ such that $$$u \lt v$$$ and $$$w_{uv} \lt \infty$$$, we want to check if there exists a pair of vertices $$$(a, b)$$$ such that $$$d_{au} + w_{uv} + d_{vb} \leq l_{ab}$$$. This is equivalent to $$${}^\exists a,\; d_{au} + w_{uv} \leq \max_b ( l_{ab} - d_{vb} )$$$. We can precalculate $$$f_{av} := \displaystyle \max_b ( l_{ab} - d_{vb} )$$$ in $$$O(n^3)$$$ time, then check $$${}^\exists a,\; d_{au} + w_{uv} \leq f_{av}$$$ in $$$O(n)$$$ time for each $$$(u, v)$$$. Therefore the problem has been solved in a total of $$$O(n^3)$$$ time.
how sad it is when codeforces does not work:(
Rounds for children are so bizarre, each time I participate in them it's such a disaster.
The mice cried, pricked, but continued gnawing the cactus...
From Russian saying.
Yes, it's from the anecdote where the mice wanted to become hedgehogs, or alternatively from the story in which a real cactus was hurt by hungry mice which weren't able to find some less painful comestibles.
Is this round rated?
The problem E is a very great problem, especially limited the number of inquiries <= 10^{14} and let my multiplication get TLE(because of invalid queries) Thank you very much!
B > C. Just another day at CF.
Can't wait for editorial to see the intended solution for Div2 D. I couldn't do it in less than O(n^2)
I have ~ O(n) solution using vector of deques
would you mind sharing the idea behind it?
Let's consider sequence p[1], p[2], ..., p[k], where GCD(p[i — 1], p[i]) != 1 for each neighbor pairs. We can store this sequence as a deque. Let's simulate the process and parse the given array into these deques.
Then we have obvious property: if you delete p[1], you start from p[2] and go until p[k], and for each neighbor pairs GCD is also > 1. We don't need to check it twice, let's just remove the first element from the deque.
After that let's consider two neighbor deques q and p. If GCD(q.back(), p.front()) != 1, we can merge them(you can use small-to-large to do it more efficiently).
You can do it while it is possible to remove at least one element. After few operations we'll get one deque, let's consider it as given array and do the same.
That's neat, I regret not being able to solve it
geranazavr555 when will codeforces be fully open ?
After the announcement of the results of the technocup is over.
Why is viewing user profiles blocked?
I guess because of techno Cup
And the rationale for this is what?
Maybe because you can look at their solutions since it was a "different" contest.
To avoid leakage of technocup results
Can anyone give any hints for Div2 C?
In Div2 C why it was not given that "nobody is chosen strictly more than " Ceil of m/2...I got wrong thrice on it due to assuming it to be rounded down :(
I also get confused each time.. Remember in ceil, brackets are curved at top and in floor, they are curved at bottom..
It was mentioned
My soln of D -
Let d[u][v] be the shortest distance between u and v.
O(n^4) soln - Let x,y,w be an edge. u,v,l is a triplet.
x,y is good if
d[x][u]+w+d[v][y]<=l<=>0+w+d[v][y]<=l-d[x][u].Its equivalent to adding (x,v,l-d[x][u]) as new triplet.
You can always remove duplicate triples by keeping the one with has l maximum.
So you can use initial given triplets to generate O(n^2) new triplets.
At last, for each edges just check triplets which have an endpoint common with the edge.
Overall it takes O(n^3).
Did someone else use flows on Div2C??
I did that too. I'm surprised so many people had greedy instead of this solution.
how did you saw other people's solution ..I'm not able to see anyone's solution from standings??(its just loading..)
From discussing with other people.
Yeah! Dinic just worked fine.
Was reluctant initially to implement as was quite sure it will TLE. Implemented it when the round was about to be over. Funny.
Dinic on bipartite graphs works with complexity $$$O(|E|\cdot\sqrt{|V|})$$$.
Yeah, I'm suprised that the Greedy apprroach work in this problem, my friends say that they use Greedy algo to solve without proof = )
Really without proof ???
A was so easy, everyone signed up for the contest, but the surprise was B. oh shit, contest is not beautiful
hello! it's beautiful. listen to me.
This really makes me hate myself.
Oh captain，my captain！
div2 C problem: if he has only one friend, how can make other friends offended?
lol , nice one.
Any hints for div2C?
one more
Greedy works.
HINT 1:-
if all friends appear in less than ceil(m/2) ... ans is pretty simple
HINT 2:-
if any friend(let's call him x) appear more than ceil(m/2) times then invite this friend for ceil(m/2) times on days and for rest days invite anyone(except the one you invited ceil(m/2) times already). This will automatically obey the condition.
Now on which days should invite x.
Invite x on those days in which only he is allowed(let call this num y) and rest on any day in which x is allowed.
if(y>ceil(m/2) then ans = "NO" ..... It is obvious.
I hope this helps
I will try to explain what I did... Firstly I choose those friends whose count in all games (i.e-in all m days) is less than max allowed, Max allowed is ceil(m/2).We can use these friends on all days wherever they are possible because they can all be used at once and it will always be optimal and it will not violate the maximum usage condition. Now you can now take all the friends whose count is greater than max allowed.
Now the question arises which friend to use and on which days to use. So, for each friend you can choose the days where there are less possibilities and that particular friend is possible on that day.I hope you understand.
Sorry for my poor grammar.
110670278 Can anybody tell me what is wrong in this code. Thank You
I pass pretests in C, but why it doesn't turn green nor turn red?
Does it mean FST?
I had same issue with div2A. I got this reply : Some features may be disabled due to hosting Technocup, sorry for that. You got AC for A
Check the standings. For some reason, the problems page is bugged.
Same happened with me for A, when CF was down my same solution was submitted twice, due to which got 50 points penalty ;(
bring back the good old codeforces!!!
Why haven't system tests started yet?
But why does Div2 D/ Div 1 B have to be literal dog shit? It's almost like the people who proposed to let this kind of garbage appear in a Codeforces round don't want anyone to have fun participating it.
+++
Could you please explain why it was a bad problem ?
any hints for ques div2-B ?
I have some serious doubts about performance of alan4ik. I don't believe that any purple user is capable of solving B and C significantly faster than all reds (he solved B at 0:03, A at 0:07 and C at 0:16) and this competition being a mirror of Technocup explains pretty well how it was possible
some users in div.2 solved first 3 problems in starting 1 minute of the contest
I Encountered some problem while trying to submit problem A, which led to a delay in submit time and a down my ranking in the standing list, and the resulting from it confusion prevented me from focusing on the rest of the issues. Which of you happened to him like this?
Please add (n = 1) case to pretest in next contest. TY.
Me after FSTing on B : Hey cyan, here I come !
The rating predictor doesn't show anything for me. Strange!
What does FST stands for?
.
Weak pretests are so common now in codeforces.
in second test case of div2-B i am getting this result- ''wrong answer Jury has answer to test 34, but participant has not (test case 34)'' what does this mean ?
you must be printing 0 or -1.
So painful to solve D just three minutes after the end
Yet so lovely that so many people got FST in B
I solved D on 2:14:58
Is your solution a Bruteforce ? What is its time complexity ?
Whenever I delete some index at most I recalculate for 2 value because things change just for this index's next and before indexes currently in the array so I recalculate them.
We can delete at most N indexes.
So total complexity is O(N) but I am also using set hence my solution is O(NlogN) but can improve to O(N)
Okay thank you ! Got it !
I see that the blog is being downvoted. What was wrong with the problems? I've heard that in the statement of D it was unclear whether the pairs are ordered or unordered, but I don't know what else was the fault of authors.
this is at least disrespectful to the author
D2B FSTs is the main reason I guess
My suspicion is that this is largely motivated by the first two Div 2 tasks--A is significantly simpler than past Div 2 A's, and there were a number of ways to attempt B that led to lots of nasty edge cases. That said, I thought the Div 1 set was pretty nicely balanced. It's perhaps a little unfortunate that the range of problems that matters most for the majority of contestants (B/C/D) leaned towards the DS side of things, with the ad hocs placed at A and E, but that doesn't seem like the end of the world to me, and neither B nor C required absurdly nasty implementation.
Another plausible explanation is that there were a large number of FSTs. I see this as perhaps a more legitimate reason to downvote; FSTing is extremely frustrating and unfair, given that it represents a significant disadvantage over failing pretests in spite of the fact that the root cause of failing pretests and of FSTing is essentially the same (i.e., submitting a buggy solution), and consequentially, I think authors have an obligation to minimize the number of FSTs that occur.
As far as I check all the problems either in div1/div2 had some people FSTed so I think that's the reason.
Huh? That's actually great. For a long time cf pretests are going in the wrong way. People stopped even caring about their solutions after seeing that the pretests are passed, while they should carefully check their time, memory and time relative to other participants. Pretests are here to help you catch bugs, but you should always be aware of on WA/TLE/RE on systests and be ready to resubmit if your running time on pretests isn't good enough.
Last times I've seen that after systests my running time sometimes decreases which means that most times all the strongest tests are included in the pretests, which is bad.
Isn't the point to construct as strong pretests as possible? I think pretests are here just to reduce the load on the judging servers, not to force participants to reread / test their solution after passing pretests.
Actually I've never thought about pretests from this points of view, but probably this is the actual reason.
Anyway, for me it always was another interesting feature of cf contests in comparison to other platforms, so I would've liked more tasks with weak pretests to make competition more interesting.
Is the contest still rated? please, let me know.
how n=1 is not in the pretest?
it literally took me 1 minute to detect my error :(
is CF-predictor working ??
Nope. You can't visit other's ids. Maybe it is the case for the api as well. Cf tool must have stopped too i guess.
any reason for that
I am not exactly sure how cf-predictor works but it should have to fetch some data for your account. Since visiting ids was blocked by cf till some moments ago, cf-predictor was not working. It should be working now though.
мое решение получило ТЛ 20 на систестах, а переотправленное уже прошло. Посылка которая получила ТЛ 110676013, и посылка которая прошла 110673324 KAN MikeMirzayanov
Bro, you're orange and you don't understand that there are natural fluctuantions in tests time and the consequences of that, seriously?
he was red hello and you r blue, think u r smarter, seriously?
You don’t understand that if solution gets TLE, then it is runned several times and if any of those runs gets OK, then your solution is considered correct. Seriously?
I understand that it is perfectly natural because of fluctuations
It would be one thing if the same solution passed with half the time of time limit
But 4960 out of 5000 — this is ridiculous
If your solution is on the borderline then depending on your luck it may pass and may not.
AFAIK even now in case of TLE the code is restarted two times to mitigate the issue. But there is no way to solve the issue completely.
110676013, успокоился? хватит умничать
What should be output for this TC in Div2B 3 4 2 0
-1
Is this same case ? 3 829404334 757138662 684872990
nevermind i was wrong
if n=3
4 2 0 should give 2 4
c should be less than m
Edit — if opp case then arr[1]!=m
actually .. my first comment was right it shoud give -1 as for the second case it should give -1 too
see TC 2 of div2B it is showing 0
It should be 0. c = m-2 and m can be very large
ok thanks
For div1A/div2C:
I used Dinic to solve it, modeling a graph with Source that gives (M+1)/2 to "left-side" nodes(that represents "friends"), adding an edge (of weight 1) to "right-side" nodes (that represents days) when that friend could be used on that day. Then i joined the "days" with SNK (weight 1) and run maxflow.
Anyone knows if it can work with Edmond-Karp or some Matching algorithm?
You dont need to struggle that much. Although I dont know matching algos, it can be solved greedily. First go through all the days for which there is only one friend available, take him and increment his count.Then go through all the remaining days, take any person whose count is less than ceil(m/2). If at the end, there is any person with count > ceil(m/2), then print no, otherwise print the result.
Hopcroft Karp algorithm should also fit.
Why https://mirror.codeforces.com/contest/1483/submission/110663953 gets TL, while https://mirror.codeforces.com/contest/1483/submission/110672560 gets Accepted in 4570 ms? The two files are identical
How the hell were you able to submit the same code twice? Didn't it give any error?
The second submission is in practice, it seems that the system does not compare contest submissions with practice submissions.
Well, I have no clue why there is run time difference in 2 of your submissions but I want to say that hard-luck buddy! luck wasn't on your side in the contest.
I have a doubt. What is the correct size for declaring segment tree ? For problem C (skyline), I declared N as 3e5+5 and seg tree size as 3N. But it gave memory error on test 6. Then When I changed to 4N, it worked. Since the seg tree divides every node by 2, so the max size should not go beyond 2N right ? (N + N/2 +N/4 +... = 2N) ?
Yes, but the indexes of the nodes of the segment tree are for $$$N$$$ being a power of two. Therefore, you should extend $$$N$$$ until the lowest power of two not less than $$$N$$$.
Even by that, 3N = 9 * 10^5 which is more than 2 * 4*10^5.
2^19>5*10^5, not 4*10^5, so 2*5*10^5>3N
Oh Yeah. Thanks.
you should have $$$2\cdot 2^{\lceil\log_2 n\rceil }\le 4n$$$.
Thanks.
Для участников технокубка раунд рейтинговый?
I passed div2 A at 0:08 but why did my submission is skipped？ I submitted div2 A again after i passed the B，but my A‘s score is not calculateed with the submission i submitted at 0:08。my score is lost！
It wasn't lost. The dashboard was faulty. It didn't show green for A. But it was in the submission page. Then you submitted again. Therefore your previous submission is skipped.
How to see the probable rating change before the end of the round?
To not keep you waiting, the ratings updated preliminarily. We will remove cheaters and update the ratings again soon!
So,will the rating changes be rolled back ?
Finally.... reached expert today :)
Congrats man.
Thanks:)
Now you can get your girlfriends https://mirror.codeforces.com/blog/entry/84657
Congratulations bro ;)
Thanks :)
in Proplem B Div 2
When can I say that the condition takes a value of 0?
In three cases:
1) whole array is increasing (with the same difference between adjacent values)(like: 1,3,5,7)
2) whole array consists only one integer (like: 1,1,1,1)
3) whole array is decreasing (with the same difference between adjacent values)(like: 7,5,3,1)
if the difference is not same in case 1 and 3 then answer will be -1
Thanx
when will ratings be updated ?
Already been updated!
anyone getting notifications that someone wrote something ?
Yes(if writer is my friend).
Some of my friends wrote comments and I didn't get anything
https://mirror.codeforces.com/blog/entry/88849?#comment-773057
div 1.5 i guess
How to solve Div2D / Div1B?
My solution is following. Notice that we will remove at most N numbers. So, if we can "fast" search next and remove it, then time complexity is O(n * (single_search+single_remove)). How can we remove element? You can do it with double-linked list or using set. How to find next pair? Let's maintain set of all indices of consecutive gcd = 1. Then, we need to find among all indices next from current position. In c++ you can do it with set and lower_bound. Or you could write segment tree. Probably there some other ways. When you delete number make sure you also remove all corresponding gcd = 1, and add new ones.
Thanks! Got it.
In DIV2 A and B, what arguments or justification you used while reaching to the solution ? Like what was your lane of thought .
In Div2A we can think of a cell as a vertex in a graph and a broken wall as an edge. Let's also assume that there's a vertex that represents "outside of the prison". Then, the task is to find the minimum number of edges that a connected graph can have. In other words, the graph we're looking for is a tree which has n-1 edges. Including our "imaginary" vertex (because there must exist a connection with the "outside"),
n = a * b + 1so the number of edges isa * b.A: You can think of (a*b) cells + 1 outside as components of a disjoint set. Opening any wall means joining two componenets. As we need to merge a*b + 1 components to 1 we need at least a*b wall breaks. Then you can easily create a solution which satisfies this for eg. breaking the bottom wall of each cell.
B: If the array satisifies a generator there can be only two cases for each pair of adjacent elements: a[i] >= a[i-1] and a[i] = a[i-1] + c, and a[i] < a[i-1] and a[i] = (a[i-1] + c — m).
After checking that each case has only one difference in it, you need both these cases occuring to fix one (m,c) pair else m can be arbitrarily large.
Getting error 'You are not allowed to view the contest' when trying to open Div2 contest (ID 1484). Please fix, I'd like to look at the problem statements again.
MikeMirzayanov KAN
I found out that the link to the div 2 round is not working. It said that I am not allowed to view such contest! Please fix it. @MikeMirzayanov
When can we view others' code?
It seems the prompt for Div2 Problem B was misleading. It said that c is nonnegative and that a[i] = a[i-1] + c. That led me to conclude that arrays that are monotonically decreasing have no solution, but that doesn't seem to be the case based on some of the comments (Codeforces is not currently allowing access to the div2 problems so I can't check the testcases I missed).
a[i] = (a[i-1] + c) % m. Not necessarily monotonically increasing or decreasing.
But if the array is monotonically decreasing by the same amount, the mod operation wouldn't change anything would it? This array for example: 10 8 6 4 2
Using the terminology in the statement (https://mirror.codeforces.com/contest/1484/problem/B), n=5, m=1000, c=998, and s = 10 works
Codeforces Round #709 (Div.1 only)
Is there a reason why problems A,B in Div.2 cannot be viewed?
Can anyone please explain how to solve DIV2 C (DIV1 A) using flow?
The problem can be reduced to finding the maximum bipartite matching of the friends with the days, restricting the capacity from the source to each friend to (m+1)/2.
Dude, why do you implement flows to solve this simple problem? Anyways using flows will most probably give TLE.
I'm not sure if greedy will work if constraints were changed to m/4 or m/3. Also I cannot prove if greedy is always correct, it's just seems intuitively correct because of large constraints.
Greedy works because after any assignment there can only be one invalid friend. Then while greedily reassigning friend during the days where this invalid friend was chosen, there is no possibility that a newly assigned friend will become invalid, because if we consider the frequencies of the previously assigned and newly assigned friends:
For even m closest case would be: $$$(m/2 + 1,m/2 - 1)$$$ to $$$(m/2,m/2)$$$
For odd m closest case would be: $$$((m+1)/2 + 1, (m+1)/2 - 2)$$$ to $$$((m+1)/2, (m+1)/2 - 1)$$$.
In both cases, the invalid friend becomes valid and the valid one stays valid. Of course it woudn't work if the limit was changed, that was the point of the problem.
Yeah... makes sense, Really appreciate the effort
can we use dp, where we maintain the frequencies of the friends in a map and for each day if the frequency is less than ceil(m/2) then we push it and call the next day recursively? and we end it when the call has an empty list of list of friends ?
Hey, I just did it using the same way, but getting WA, could you please help me in telling the case I am missing. IMO I don't think you need more than two friends to come on a single day to get to the answer. Submission: https://mirror.codeforces.com/contest/1484/submission/110655268
But when it got a WA to me, so I expanded it a bit for checking for all friends. Submission: https://mirror.codeforces.com/contest/1484/submission/110660421
But I not able to find out the Error.
im getting WA on tc2 for the same reason too, even I wanted to know why dynamic programming is not working in this case.
Ig this is not always optimal. Suppose x and y are available on second last day and only x is available on last day and count of x is (ceil(m/2)-1) and count of y is 0 till third last day.
So now we should take y on second last day and x on last day for optimal answer. But as per your approach you can end up selecting x on second last day and having no choice on last day.
When I chose x and have no choice on last day , the call comes back negative for the call in which I chose x and the function proceeds to chose y and check again , this tome it will get a true
I can't access your newest submission. Try the test case commented below. Your earlier submission(which I can access) fails in that case
You should see my submission once again. I am handling the cases with only one person on one day initially. And then checking for others. There won't be any case because we will first check for the day with only one person.
1
2 3
2 1 2
1 1
1 1
Your code gives NO
But answer should be YES
MikeMirzayanov почему мы не можем смотреть коды других участников из DIV1?
These rating changes don't seem appropriate, is there some extra +ve rating given in the first few contests? , I would like to know more about this...
See this.
Would you please describe the details of the complexity analysis about Div2. D in editorial?
This problem is still puzzling me.
Where did you get the editorial, please provide the link.
I just realized I got only 438 points for problem A, but I did solve it in under 5 minutes. There are many who took more time and still got more points, is it because I submitted my problem twice? because the first time the server went down while it was in queue.
Submitting multiple times result -50points at each resubmission.
where is div2? I cant open it.
when to publish the tutorial?
can anyone please explain the approach to problem D? i have no clue how to do it..
MikeMirzayanov Please update problem rates.
MikeMirzayanov Div3 round problem ratings are updated but problem ratings for this round is still pending :|
Sorry if it's too late, but i think C div.1 can solve using divide and conquer and greedy with a bit more data structures.
See my code : 110746829
Let $$$solve(l, r)$$$ be the answer for segment $$$[l, r]$$$ then you choose the building $$$m$$$ with minimum $$$h_m$$$. You easy calculate it in O(1) if you have the answer for segment $$$[l, m - 1]$$$ and $$$[m + 1, r]$$$. Let's call these answer are $$$left$$$ and $$$right$$$.
There're two case here:
Take max all of these, and you got the final answer.
Any plans to add english editorial authors ?
KAN any updates on the english editorial?
Thanks to everyone who downvoted my survey is accomplished. Also, thanks to Dragnoid99 to make it possible faster. ORZ community
First of all, this was a useless comment and second no-one can judge whether you are dumb or not.
waves
At this point I think I will just try google translating the Russian slides instead of waiting for the editorial.
god! Am I the only rookie praying for tutorial?
Instead waiting for English tutorials to be posted, i am going to learn russian to speed up the process.
Here is the English Editorial version (using Google Translate) of "Slides in Russian" which is attached to the Announcement — Google Translate Version Of Technocup 2021 Final(CF — 709) Editorial .
Hope it helps before the official editorial gets published.
Thank You !
Sorry, I mistook.
I am not the editorialist.
I have just translated the Russian one which is attached to the Announcement using Google Translate and put them in a Document for the curious ones.
UPD : Glad you understand my point.
In case anyone is curious about the solutions to the problems, I wrote up an unofficial editorial for all the problems except Div. 1 F at https://mirror.codeforces.com/blog/entry/88944.
please post editorial asap !!
I'm not sure why editorial is not out yet, but I was the writer of Div2E/Div1C, so ill post the editorial I had written for the solution I had in mind, hope it helps out.
We can solve this problem with DP. A trivial $$$O(n^2)$$$ algorithm would look like this: Define $$$dp_i$$$ as the maximum beauty that can be achieved if we have a set of photos of buildings from $$$1$$$ to $$$i$$$. We can check every possible splitting point $$$j \le i$$$ for the rightmost picture of the set, and keep the biggest answer. $$$dp_i = max_{j \le i}(dp_{j - 1} + b_{j..i})$$$.
Now we just need to optimize this solution. Assume we are calculating $$$dp_i$$$ First important thing we need to realize is that, if we find the position of the closest smaller number to the left of $$$i$$$, on position $$$j$$$, and we choose to add it in the rightmost photo with building $$$i$$$, then the best solution would be on $$$dp_j$$$, because all numbers after $$$j$$$ are bigger than $$$h_j$$$, so they would not change the beauty of the picture (this is assuming that $$$i$$$ and $$$j$$$ are on the same photo). Note that we had already calculated the max beauty of $$$dp_j$$$, so it is not necessary to go back any further, as we have the best answer stored at $$$dp_j$$$
Having this observation, we are just left to check numbers between $$$j$$$ and $$$i$$$ as possible splitting points for the rightmost picture (the case where building $$$j$$$ and building $$$i$$$ are in different pictures). But we now know that every height from $$$j + 1$$$ to $$$i - 1$$$ is bigger than $$$h_i$$$ ( this is because $$$j$$$ is the closets smaller height), so the answer will just be $$$dp_{k - 1}$$$ + $$$b_i$$$ for any $$$k$$$ between $$$j + 1$$$ and $$$i$$$. We want to maximize the answer, so we just want to look for the max $$$dp_k$$$ value in this range. To do this, we can keep a max segment tree with dp values, and query it in $$$O(lgn)$$$ time. After we calculate $$$dp_i$$$, we insert it to the segment tree. This gives un an $$$O(n*lgn)$$$ solution, enough to solve the problem.
For the final implementation, we can iterate from 1 to $$$n$$$, keeping a stack with height values, to calculate the closest smaller building for each building. We just pop the stack while the current building is smaller than the top value of the stack, and insert the current building on top of the stack. Actually, by using this trick right, a segment tree is not really necessary. We can calculate the minimum answer for the ranges by updating information as we delete or add numbers in the stack. So it is possible to achieve a linear time solution. However, $$$O(n*lgn)$$$ is enough to solve the problem, so this optimization is not necessary.
Thomas had never seen such long delay for editorials
Please, upload the editorial. It's been 3 days.
It's frustrating to wait so long for the editorial
Can the editorial link be posted on the announcement also? KAN
How to solve div2F/div1D?