Hello Codeforces! We (dqa2021, Xiejiadong, Retired_xryjr233, Z18 and me) are excited to invite you to take part in Codeforces Round 664 (Div. 1) and Codeforces Round 664 (Div. 2), which will happen on Aug/12/2020 17:35 (Moscow time).
Huge thanks to:
- 300iq for coordinating our round.
- triple__a for answering my questions about contest preparation.
- duyiblue, jhdonghj112, jqdai0815 for providing and discussing ideas.
- gamegame, KonaeAkira, gtrhetr, jqdai0815, mayaohua2003, crasysky, aryanc403, RainAir, dqa2021, dingwanren, world_top_1_AD_uzi, Juanzhang, _Aaryan_, GavinZheng, taran_1407, yijan, whuang369, jhdonghj112 for testing and providing feedback.
- MikeMirzayanov for Codeforces and Polygon platforms.
There will be 5 problems in Div.1 round and 6 problems in Div.2 round. You'll be given 2 hours to solve them.
The story of this round is about that man. Instead of displaying his name, I prefer telling one of his legends (or joke):
"I have a 'friend', who makes lots of money every day, earning a billion in the blink of eyes. With a wave of his hand, OIers all over the world will follow him."
You can post your guesses in the comments.
UPD: Score Distribution:
- Div.1: 500 — 1000 — 1500 — 1750 — 2500
- Div.2: 500 — 750 — 1000 — 1250 — 1750 — 2250
Good luck!
UPD: Congratulations to the winners!
Div.1:
Div.2:
Auto comment: topic has been updated by sshwyR (previous revision, new revision, compare).
What's the point of declaring top 5 people in Div.2, after all they are just Div.1 guys who registered with their alternative account, isn't this unfair? for example just look at the person who is placed 3rd in Div.2, this guy was newbie, like seriously? This disrupts entire rating distribution.
I am tester, please upvote for me! :)
( I'M CUTE, GIVE ME CONTRIBUTION )
I am tester too, please upvote for me! :)
(I'M CUTER THAN duyiblue, GIVE ME CONTRIBUTION )
LOOOOOOOOL~~~~~
You are cute too.
haha
I am tester too, please upvote for me! :) (I'M THE MOST KAWAII TESTER IN THE TESTER TEAM QAQ)
How kawaii you are! Would you mind tell me what character she is?
avatar She is warship Queen Elizabeth from a video game called Azur Lane.
I am not tester, please upvote for me :) (I am cute)
i dont think tourist was there in this round.
I am tester tooo,plz upvote for me!:)
(OBVIOUSLY I'M CUTER THAN duyiblue AND jhdonghj112 QAQ)
Actually duyiblue and dingwanren are the same person?
So double contribution!
Actually they are not the same person.
Actually they are
dingwanren is a imitation of diamond_duke (an OIer in Senior 3 in NFLS) by duyiblue.
If you take a good look at duyiblue and dingwanren's programs, you will find that they have the same way of writing programs
Ah,I thought dingwanren was diamond_duke for I only see the profile page. What a mistake!
orz tzc, tzc tql ddw.
orz dy, dy tql ddw.
I'm cuter than duyiblue , I'm his girlfriend () please upvote for me
I think you should test the round next time instead of asking for contribution!
Hey tester, Why ruined the day of so many cute contestants with weak pretests?
Good job testers very good round, glad nothing went wrong!
Sorry, but As a tester, you don't do a great job! Many solutions failed(even of LGMs)
Is the man is Bill Gates? :D
very close (in their amount of money), but not.
Elon musk ?
Boboniu！
Jeff Bezos ?
is it kanye?
You are thinking it's spam that's why I removed it and going to do some practice otherwise in search of the name I will lose my ratings. XD:)
This shitty name is spamming everywhere.. Please stop that.
Edit: It's not a spam, it's a trend. Do some research.
is that man Jack Ma? just a random guess
We don't want stories please, as they come in the way of easily comprehending the problem statement. :(
Oh don't worry. The statement is not long and we just use the main character to describe it. The story itself is easy to comprehend :)
For me who is afraid of number theory, this is good news to be happy about. If there is a math problem, I am also ready to change to the alternate.
Tester — Planning to participate in the next div 1.
sshwyR — You already tested it.
Tester -
It's the round we tested 3 months ago and didn't knew when it will be scheduled. Many of us found we tested upcoming round (and hence cant participate) by looking at our names in the announcement.
MeToo
This contest was put on hold for a long time due to the difficult Di2A.
difficult Di2A
Thank you , now excuse me while I go to sleep for 35 hours real quick
Go for 40 , than that would be sacrilegious
Two set violin = small pp
Bass = pp large
What? You bass gang, had some arguments before, but now is just butthurt, VIOLIN > bass. TwoSet responded to all Davie's videos, but Davie just ignored the "bass diss track" (which was asking for a livestream at the end) and the "davie fakes playing the violin". Brett played bass poorly, but it was REAL, but Davie, on the other hand, FAKED playing it. So what side is winning, huh?
Oops, difficult Di2A. So scary....
Is this difficult div2A? Then what about the contest, which was for Russian elementary school students, with a 1700 rating problem A. (it was like 10 contests before) :D
Feeling sad for such testers but they will get contribution.
Well I knew I wont participate in one div1 when I was testing. Anyways I did participated but in unrated mode. Just that I didnt knew I wont be participating on 12 Aug.
Most testers vc contests before round.
Another Chinese Round! This round has special and interesting stories.Good luck and have fun for every contestants. Some information in advance:Those stories may all be about one mysterious person.
[O)]-[(O]/
Should i wear mask and senetizer during contest??
You can if you are comfortable
Is the man @MikeMirzayanov ? <3 i have not any guess about his income but i love him and want to see his name in the problem statement ..
no no no
Mukesh Ambani???
Boss Cai!!
WoW
boboniu!
Are you talking about me?
Wow!
I am almost sure this man is Jack Ma.
I want to know what is Polygon platform ? Strange word for me
A platform for making problems
https://polygon.codeforces.com/
Polygon in basically the platform for creation of programming contest problems. It also supports problem statement writing, test data preparing, model solutions, judging and automatic validation.
I know! The man is Cai Rui[user:Boboniu]!
Intersting fact: tourist will set a new best rating ever on codeforces if he wins this round!
Legends don't care about rating (¬_¬ )
but we care about legends rating :)
Chinese Round?
Fun fact: seeing MiFaFaOvO can't participate in this round, if he doesn't in the next global round, he will vanish from the live standings xD
And "poof": he is gone xD
Excuse my ignorance, but what is the rule exactly you're talking about?
Only the users which participated in a rated round within $$$6$$$ months will be included in the
Rating Standing.The last rated round for him took place on $$$Feb^{17th}$$$.
Oh. Didn't know that about the 6 months rule. Thanks!
I guess it's CCF
Totally agree!
Coin Collecting Federation (
I do CP for fun.. What about you guys?
I do CP for codeforces, and codeforces for the memes.
For practice.
BINOD !
Please don't spam Codeforces at least. There are many who don't know the context and it gets annoying after a while.
Alright . Apologies!. Will be careful next time.
It's Mr-Cai!!! Hahahaha
When I see Chinese round, I think of Mathforces.
hhh... don't worry, I promise it'll not gonna be Mathforces.
This round will not be math round, so please feel free to compete.
Mathforces confirmed
I love mathforces
Nope! Just FST Forces!
so... Yet another maximizing profit problems ... or Yet another guessing name problems :v
Ronan Ryan?
interestingLSY Time for your picture about boboniu!
Ah-oh, jqdai0815 as a tester, which means we can't see his competition with tourist this time. A huge pity!
Either way ,he will never compete against tourist ,after the time he gained 1st spot . He just prays for some bad luck for tourist ,so that he can remain on top.
Before that ,i Never thought ,such legendary grandmaster would ever fear to directly compete against someone. But anyway that someone tourist is not ordinary guy.
Don't comment others at will.
？？？？？
Don't comment on Lagendary Grandmaster subjectively!!!
No one would like to be commented like this
But It's still an interesting and exciting contest
Yeah
He must be Zide Du(杜子德) from the CCF.
China Computer Federation or China Collecting-money Federation?
Is him, right ?
Please provide short statements for problems and understandable english,I am learning english .Thanks
Copy to baidu translate (
Website: fanyi.baidu.com
[Deleted]
You take slander as freedom?
You should know it's not polite to comment others like that
Freedom is not your excuse for slandering.
Although it is your right to express your own ideas,but your last comment was unsuitable indeed in some way.And codeforces delete your comment for a great many people downvote you.I think it is reasonable.
See this 89567737
What's happening to "timeanddate.com"? :|
The website works well for me, probably it's because of your network problem.
And I don't think these sorts of comments are meaningful here
as#holes
Excuse me is there anything wrong?
Boss Cai -- boboniu
Orz.
Makes lots of money everyday. That must be dzd from CCF(China Collecting-money Foundation)!
I became psychopath because of the little pony, please make sure that the problems have short statements. with appreciation to your great story.
The little pony turned you psychopath?
..
I really hope that the problems are not as hard as the quiz.....
How many number of common problems ?
Richie Rich?
It could be suneo's dad from doraemon xD
is it jeff bezos ?
Olers?
Realy, what is an Olers?
Olympic of Information ers
It's what Chinese called people who participate in Olympic in Informatics contests.
Is that man Mark Zuckerberg??
Rick Sanchez with Jerryboree Idea
Is the man Mike Mirzayanov ?
JPow?
Money printer goes brrrrrhhhhhh
If there are unnecessary stories, please, do mark them in Italic or any other way. I've faced a lot of issues about bad storytelling in Codeforces. :)
What are "Olers"?
The people like us. Take part in OI. OI, Olympics Information. OIer, the people learn OI. OIers is plural form of OIer. Understand? :)
I'm not good at English. I hope you can understand xD
OIers are coders who participate in programming competitions (mostly used in China I guess). OI = Olympiad Informatics
It's what Chinese called people who participate in Olympic in Informatics contests.
The goalkeeper of RCD Espanyol, click the link for more details.
https://www.transfermarkt.de/oier-olazabal/profil/spieler/8176
Since Wu Lei (the best football player in China) is also in this team, we use the word "Oier" to show our respect to RCD Espanyol so that they pay Wu Lei on time.
i guess Jack Ma .
I guess he's kkksc03, head of Luogu(an oj in China). For every monthly contest in Luogu, if you want to watch the video tutorial, you must pay 10￥ first. And what's more there's lots of OI online lesson made by Luogu in every summer and they're not free! So he can make lots of money in this way.
Actually, according to himself, the money is not paid to him directly and a lot of money is spent in maintaining Luogu's website. And the lessons' money will be given to the teachers. So I'm very sure that it's not him.
Any hint about no of common problems or score distribution ?
Score Distribution?
Obviously,the man is DUZIDE:P
Score Distribution?
Sorry for the late publishment of score distribution :)
I have a request for MikeMirzayanov.. can we have TEAM contest at codeforces.
Team contests are a nice idea, CodeChef already have it though, it's called "LONG challenge"
Long challenge isn't a team challenge
I agree. XD
It's €€£
sshwyR how many problems will be common in both divisions?
I guess there are 4.
D2A->D2B->D1A/D2C->D1B/D2D->D1C/D2E->D1D/D2F->D1E
upd: When I saw the scoreboard I knew there are only 3 but I forgot to update QAQ
3 D1A=D2D,D1B=D2E,D1C=D2F
3
Thank you!
Auto comment: topic has been updated by sshwyR (previous revision, new revision, compare).
Today's tester busy with increasing their contribution .
Div2 Problem E with 1750 (^_^)...it should be solvable
obviously rating < 1900, i guess.
E was a bit difficult :/
iam not a tester, but please upvote for my pp :).......
i see ,no feature to upvote pp ? that's why you can't upvote it :( leel
I upvote
The man must be, President of IOI: Richard Forster
Zhengru IOI will be internationalized.
Karuna! :D
An off-topic question: how can I view the past contests right before contests?
Change the url. It will by default have contestId of the present contest, just remove that and you will be good to go.
When I access to https://codeforces.me/contests/ , it automatically redirects to https://codeforces.me/contests/1394,1395 .
There is an asterisk sign below the upcoming contest tab that states , that to view the complete list click here.
Ah, that was the link I didn't know! Thanks a lot.
But when you are already on https://codeforces.me/contests/1394,1395, you can just delete from the link 1394,1395. For me it works. And of course you can use method mentioned by absr007.
its boboniu
Too much if else... :(
Normal Div1:
Try to solve more problems
Horrible Round #664 Div1:
Try to solve the first problem faster
No,you should try to avoid FST instead of solving problems faster. It's my final standing:
(I was rk128 before the system test...)
If I accept no question am i still rated?
why they set so hard
Was this the "Single Test Case" Contest?
Really nice problems and clear statements. Enjoyed taking this contest :D
Codeforces rounds are great, they teach us how to look at a problem in a simple manner, without overcomplicating it. orz Great contest.
Well well well, once again someone posted solutions on youtube during the contest, you can even see his username in the video, i wonder if these guys are shameless or what Video link Username: Khayrulmithu
I don't know why but that video sounds exactly like an airline pilot's announcement.
And he got it wrong on test 10 :D
I really liked d1 B, but...
Starting from any vertex u
This would be much more understandable if it said from all vertices. Wasted lots of time solving the wrong problem.
That could confuse people into thinking that it's possible to start from multiple vertices simultaneously, I think the clearest is "for all vertices u, starting from u, ..."
I'm pretty sure that "any" was actually used correctly here
What was the point of swapping X and Y in Div2B?
I got -4 because of that
Same thing bro, same thing.
Ikr, I got 2 WAs cause of that, had to rewrite it all.
Does this works for Div-2 E — Make a graph $$$g[i][j][k]$$$ = how many edges exist such that its head is on node with $$$i$$$ outgoing edges and tail is in node with $$$j$$$ outgoing edges and it is $$$k'th$$$ smallest one. Then we brute force every $$$c_i$$$. That will be $$$O(k!k^2)$$$
The whole round is like an ingenuous play of MiFaFaOvO to get back his first place in the rating (though unsuccessful). Chinese people are really good at the art of war. My respect.
(please don't take this very seriously)
"There are two ways to go higher, self improvement, and sabotaging literally the entire community"
He has made his choice.
(Jk no offense pls :)
Seems his plan worked
Why the "non-empty" in C? I guess it has to do with printing the output but it reduced problem quality by like 30%.
could u plz xplain how to solve todays div2 C problem
You do realize that I haven't read Div 2C?
ok..if someone other can help..bcz..i am very curious about it..that why my logic was wrong..even on sample test cases ..i know i was completely wrong.can not wait till editorial..xD
rank 1 guys code is very clean to understand.
Might as well allow output of negative length to boost problem quality?
Empty strings make sense, strings of negative length don't.
OH SHIT I missed it. Even my starting answer is an empty string. Luckily my code goes for large strings first when improving the cost in binsearch, so I got AC.
Is C checking if a cube like shape fits all of the given points by binary search / math?
Yes!But I am too weak at geometry,I come up with this idea quickly but don't know how to compute the area covered by several polygon...
Maintain the minimum and maximum values of $$$N$$$, $$$B$$$, and $$$N - B$$$.
Yes,but does it turn into a geometry problem??
Depends on your definition of "geometry" I guess. I didn't really use any "geometric" techniques. But yeah, you can visualize the things as hexagons.
No need to compute the area. Just find a point inside it.
Yes,But how to compute the point inside it?
Several polygons? I think it reduces to finding the minimum side length of the square in a shape like this that can fit all the points, which should be fairly easy. But I spent most of the contest thinking I had to minimize the sum of distances instead of the maximum, so I couldn't implement it in time.
Rather diamond-like (with top-left and bottom-right edges cut). Math seems awful there.
How to solve Div2C? Also, where does the N^2 check all pairs approach go incorrect?
Final Answer can be at most $$$2 ^ {9}$$$, so just brute force for every possible value, and pick the smallest.
Thanks.
Do you think there a polynomial-time solution in case final answer is large?
I have O(n^2 log m), where m is the size of the largest input integer.
Figure out how many leading zeroes we can get by picking greedily the pairs where the bitwise-and has the most leading zeroes, using O(n^2) time. The next bit has to be a 1 in any solution, so we can add that bit the the answer. Since the bit will be a 1 in any case, we can ignore is from now on: we set it to zero and find out again how many leading zeroes we can now get. Repeat until we can get all zeroes.
https://codeforces.me/contest/1395/submission/89696109
Your implementation is too elegant :)
Now I feel like an idiot,spend half an hour in implementing an dp solution..:/
Hey can we solve problem C it using Dp??
I tried some approach but its giving wrong ans.
Can you explain this approach?
Very clean aproach, fix the answer and check there are N pairs a[i]&b[j] that has the same bit positions. I spend 1 hour until I reach my dp aproach...
Is 1C a geometry problem? I thought it just need to binary_search and compute the area covered by several polygon..
I passed the pretests in Div2B with a backtracking solution, how is that a thing?
Hints for $$$Div 2 E$$$ Please?
Assume you have selected a tuple. Of course, you can now reduce the initial graph to one such that each node has out-degree of 1. Realise that for each node to be in a cycle in this resulting graph (and thus be possible to reach back to it in finite time), each node must also have an in-degree of 1.
For each node (let it be Q), traverse over all nodes that point to it, let's call these set of nodes P. You can easily determine that for which value in the tuple will these nodes in P be pointing to Q. Also, this means that these values can not co exist in the tuple, since that would mean that multiple nodes point to Q in the resulting graph. Thus, you make a list of all these invalid combinations.
Iterate over all K! tuples, and for each check if they consist of an invalid combination. If not, then it is valid.
Sadly, was a bit late while implementing in contest :(
Here's my AC code: https://codeforces.me/contest/1395/submission/89731507
What's the upper bound of the number of those invalid combinations?
By an invalid combination, I mean pairs $$$((i, j), (k, l))$$$ such that value $$$j$$$ at index $$$i$$$ in the tuple, and value $$$k$$$ at index $$$l$$$ in the tuple, are incompatible. So the maximum number of such invalid combinations are $$$K!(K!+1) / 2$$$.
You just have to store all these pairs in a $$$K*K*K*K$$$ matrix.
Thanks a lot.
1250 not enough points for Div.2D
Long Long in problem A just ruined my whole contest.I almost wasted 50min behind it.
In Div2C, Sample test 3, what combinations get the final result as 147? The minimum I could get was 177.
I think you applied the dp approach. I did the same thing at first and waste a lot of time. just brute force the answer
I think this was the combination
a b
0 2
1 2
2 2
3 2
4 3
5 4
6 1
7 2
these are indices of the element from array a and b
I also got the same error during the contest but now I understand. Well, let's say our solution consists of two parts, 1 and 2. I am not going to explain the second part assuming you already know it.
But in the first part, u must be choosing a minimum c1 among all possible c1 and then doing part 2 of the solution but we are doing a mistake here. Actually, part 1 of the solution has n possibilities. Choosing minimum c1 is just one of the possibilities. A second possibility, for example, can be choosing a minimum c2 and then doing part 2 of the solution. Similarily choosing a minimum c3 among all possible c3 can be the third possibility and so on. So there are n possibilities.
So the solution is O(n^3) and not O(n^2). wicardobeth
i will fst in d,One detail was not considered,shit.
That was implementationforces, I like that. Should be enough to become blue again :)
What was the solution to D?
Greedy.
Edit, more understandable: We devide $$$a[]$$$ into set $$$a[]$$$ and $$$b[]$$$, where all $$$b[i]<=m$$$ and all $$$a[i]>m$$$
Sort both sets biggest element first.
Then choose consecutevly to use i elements from a[]. Foreach i we can calculate how much elements from b[] can be used then, and using prefix-sum calc in O(1) the funfactor of that i.
If only I had two more minutes to fix my implementation
Figured that, but couldn't implement. Thanks, will wait till I can see your submission.
I am used to see you in blue ;)
How to solve div2 D?
1) Make 2 lists: one of all greater than m (let us call it gr), second all less than m(let us call it sm). 2) Sort gr reverse and sm normally. 3) Now compare highest unused of gr with sum of d consecutive of sm. If gr highest is greater than simply add that to ans(in other words muzzle d smallest elements of sm). Take care of these things: 1) Do the above only till you have atleast 1 element left in gr, cause you can add it in the end. 2) If you encounter condition like less than d elements remaining in sm, then dont compare with gr greatest element as you can add all remaining along with greatest, if greatest goes in the end. 3) In the end try to put all remaining elements of gr in the end of the permutation seperated by d elements each. For details see the code: 89725326 If I fail system test ignore this comment :D
thanks!
https://codeforces.me/contest/1395/submission/89731791 I also came up with similar approach but got WA on test 16. Seems like there is something we are missing.
Can anyone please tell me what's wrong in my code? Problem A https://ide.codingblocks.com/s/308173
try 0 0 2 2
Thanx a lot
Can someone tell me why this solution to 1C is wrong?89722745
I can't find the mistake :(
Oh now I find it
I forgot one line
if(zmin>zmax) return 0;:(Boboniu stands nowhere in front of Binod
Any Idea of Test 16, Div1 B ?
How to solve Div-1 B??
Need help, why did my solution for Div2B did not pass ? I have used dfs- https://codeforces.me/contest/1395/submission/89714645
.
Problem A: What is the problem with my this python code?
I decremented min(r, g, b) from r , g, b. Then count how many of them are odd from r, g, b, w. If number of odds > 1 the answer is "No", else "Yes".
Sorry for bad English.
What if input is 2 5 7 3. Now if u decrement r g b by 2, then r b g will be 0 3 5 and w will be 9.
No.of odds will be 3 (0 3 5 9)..so it return "No"
But if I remove only 1 from r g b then r g b will be 1 4 6 and w will be 6. (1 4 6 6), we can form a palindrome.
Decrementing min(r,g,b) is wrong step.
Can someone give a small hint for 'B'?
As far as I understand:
1) The requirement "should be able to return back" is equivalent to:
= edges that satisfy Ci-rules should form loops
= every vertex should be a part of a loop
= every vertex should have an active edge coming in — this is what we need to check
2) k <= 9 so 1*2*3*4*5*6*7*8*9 = 362880 total sequences of Ci.
If we were able to somehow check whether a specific sequence is good fast, than we could just enumerate all sequences and count the good ones.
Am I correct with (1) and (2)? If no, a small hint would be appreciated.
Isn't the total number of sequences = k^9?
No, since c_i <= i, not c_i <= k
No, because 1 <= C[i] <= i
Remember, C[i] — is the index of the edge you would take if there are 'i' outcoming edges. This index can't be say 7 if there are only 3 outbound edges.
Sufficient and necessary condition is that each vertex has exactly one incoming edge. Everything else in your
(1)follows.precomputation which pairs of (i, c_i) are incompatible, because they lead to some vertex having two incoming edges
P.S. It's pretty standard Markdown, so use two spaces before newline
like this
How do you prove that?
Necessary is obvious — no incoming edge, no cycle for that vertex.
Sufficient: start from vertex and start moving. At some point you will have to return to starting vertex.
If a node $$$x$$$ has no incoming edge, it's impossible to reach $$$x$$$ starting from $$$x$$$.
If a node $$$x$$$ has indegree $$$\geq 2$$$, let $$$a, b$$$ be two nodes such that edges $$$(a, x)$$$ and $$$(b, x)$$$ exist. But if you start from $$$x$$$ you can't reach both $$$a$$$ and $$$b$$$ (wlog, if you reach $$$a$$$, you return to $$$x$$$ and you haven't visited $$$b$$$); in this case, if you start from $$$b$$$ you reach $$$x$$$ at the next step and you can't go back to $$$b$$$.
what if there is an edge (a, b) along with (a, x) and (b, x) ? Then can't it be the case such that: x -> ......... -> a -> b -> x. ?
There can't be both (a,b) and (a,x)
Then how to check if some node has no incoming edge?
Each assignment $$$c_i = j$$$ will guarantee incoming edges for some number of vertices. If you sum this up over all assignments you make, just check that it adds up to $$$n$$$.
The issue is that this can double count, however note that each vertex should have exactly one incoming edge, not more, so each assignment $$$c_i = j$$$ will exclude some other list of assignments $$${c_{i'} = j'}$$$. Just make sure to never make two conflicting assignments like this.
You have
nedges, so it's enough to check that no vertex has two incoming edges. Precompute 'bad' pairs(i, c_i),(j, c_j), such that some vertex will have two incoming edges.Simple bruteforce after that.
You are correct with both (1) and (2).
Here's a good hint. Imagine that for a certain tuple of C, we get the edges 1-2,2-1,3-4,4-3. This is equivalent to saying that nodes [1,2,3,4]->[2,1,3,4] which is a permutation! How can you use this insight to solve the problem in general?
Agnimandur kilotaras thanks! Solved!
check that each vertex has no more than one incoming edge.
This will automatically guarantee that each vertex has exactly one incoming edge.
And I've spent the contest trying to figure out how to check if each vertex has at least one incoming edge. At the end, turns out I needed to check the exact opposite.
Though I looked at other solutions I still try to figure out how other people solved it without using 4d array of pair-incompatibility [9][9][9][9].
I thought the exact same thing and could not determine how to check every possible sequence in an optimized way.
Does anyone know if there is a testcase on Div1B which breaks the hashing solution for the MOD 1e9+7?
Very nice problemset, reminded me to practice.
[deleting this because it came out wrong. I'm referring to Chinas efforts in testing all of Wuhan, which is amazing, not them experimenting and creating the virus (which I think is false).
I can't delete this, so if some CF mod could for me that'd be great.
the use of "muzzle" in problem A felt rather odd.
tourist RIP rating
FAILED TEST 31 IN C? GOING BACK TO DIV4.
I am surprised by the number of high-rated coders who do not realize that in Div1B, the in-degree of each node can be up to O(n), and iterating through pairs of nodes which go into a particular node can take quadratic time in the worst case.
I am also surprised that there was no pretest against such solutions. But I think this is fair. Contestants are responsible for ensuring that their solutions' time complexities are acceptable.
Pretests were way too weak, just look at the amount of failed submissions in Div1 ...
My finish in Div1 went from 556 to 366 because of failed submissions :D
What the hell were you guys thinking. I don't usually get mad, but now I am mad. This contest honestly has made me not want to do cf anymore
lol test 31 is ruining everyone's day ig
Absolutely disappointed
my disappointment is immeasurable and my day is ruined
My only question is why?
Weak Pretests :(
WHYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY
You could've said all of this in 1 comment?
Good point
AYAYAYAYAAYAYAYAYAAAYAYAYAYA AYA AYAY A YA YA YA PSY PSY PSY PSY DUCKDUCKDUCKDUCKKK
So many main tests failing
250+ Submissions failed TEST 31 of DIV2 C FST Forces!
850+
My rank literally jumped from 1700 to 1300
Me from 3200 to 12000+ SO GOOD to know!
Worse pretests ever.
89726386 here is my solution for D, it for some reason gave no output nor any errors. its probably a silly mistake, but i cant find it. could anyone help please?
how do i make the submission number into a link? -_-
I got you: https://codeforces.me/contest/1395/submission/89726386
Also it says time limit exceeded so it probably means you have an infinite loop
yea i checked that but it isnt printing anything even if i cout before i start writing the code :/
use
endlto flush the output.I was so happy after giving the round but Wrong answer on test 31.
Going to Specialist~~~
Even though the round was great with short (and to the point although a bit unclear initially) problem statements, I found the pretests to be weak..lots of solutions are failing the system tests.
Very weak pretests for Div2 D , even the most simple side cases werent inside the tests.
literally missed one case: if the elements are all > m..... got it accepted instantly after
Yeah mine failed on tc 25 for same reason.
Am I the only one who solved Div2 C with dp? 89700923
nice one! this approach didnt cross my mind during the competition
Hey It's really a nice approach, Can You please explain it ?
$$$dp[i][x]$$$ is the value of the assingment that minimizes $$$c_1 | c_2 | \ldots | c_n | x$$$. Then, we have
Really COOL!
I did something similar but without memoisation: [submission:89706416]https://codeforces.me/contest/1395/submission/89706416
( ཀ ʖ̯ ཀ):
Alot of solutions are getting rejected after passing pretests i think
Why doing nested ternary search is correct in problem C?
Why is brute force O(n*m) failing for div2 C ?
No it doesn't. My O(2^9 * n * m) worked fine.
I can't see a div2 C submission on your profile. Can you share your code
It's not always optimal to take $$$min(a_i$$$ & $$$b_j)$$$. Which bits you are taking matters.
can you elaborate with an example please?
Let's say the answer has a particular bit set to 1, so it means that atleast one of c1,c2...cn will definitely have that particular bit set to 1. In that case we don't need to try and set that bit to 0 for other values in c1,c2...cn by doing the AND operation since the answer would still have that bit set to 1. Guess that makes it clear why taking the minimum won't help here, it's the bits you are taking that matter.
Got it, thanks for your efforts :)
Your code's output is $$$5$$$. But the answer is
(13 & 4) | (5 & 4) | (8 & 4) = 4.When you're feeling good after a contest getting A B and C, and it's looking like you're going to gain a lot of rating, and it just all gets crushed by failed system testing due to weak pretests :(
and weak implementation
Yes that is true. I just wish that I knew I got it wrong during the contest.
PLEASE MAKE PRETESTS STRONG !
My div2A failed test 10 ,and div2B failed test 39 :( Why these tests were not used in contest ?
Everyone to their before system testing score: Those son of bitch lied to me!
in B div1 I don't get why the 9!*9*9 ~= 3e7 should not pass the tests but with some optimizations it should pass. the 1s time limit is so stupidly chosen.
Your code seems to have complexity not 9!*9*9, but way more. If there is a vertex, with $$$O(n)$$$ incoming edges it will be quadratic.
Ow right. but it need just a sort and unique and it will be ok. pretest must include these kind of tests i think.
Pretests just ruined this (I don't think it is supposed to be that many "failed on systests"). Your contest suck!
Over 850+ System test failed in problem C. Including me :(
The following submission https://codeforces.me/contest/1394/submission/89716136 gives runtime error on pretest 3 on the judge server but it works fine on my local pc and on ideone when tested now. Can anyone tell me why this is going wrong? Is it because codeforces uses the CLANG compiler?
It will give runtime error when less.size() == 1, because you are accessing
less.size() - 2 (-1)but unsigned int can't be negative so it becomes UINT_MAX.The pretests were kinda weak:((
I think it's not just that getting faild on sample tests decrease our point, because sample tests are for testing our code. In this contest, I get 2 or 3 WA's on pretest 2 of problem A that is a sample test, and it decreased huge points from my solution.
I did Div2 C like this: https://codeforces.me/contest/1395/submission/89706416 with O(n^2) complexity but I'm not able to think of a proof exactly for why this is right, can someone help me out?
I too had solved this problem in the exact same way and I'm currently unable to guess why this is working. https://codeforces.me/contest/1395/submission/89730303
In fact I also tried another small tweak in this algo but this submission passed all the tests too! https://codeforces.me/contest/1395/submission/89734201
I wonder if the problem lies in the testcases
In the first loop, $$$mn$$$ is the minimum $$$a_i$$$ & $$$b_j$$$ for each $$$i$$$ and $$$mx$$$ is the maximum $$$mn$$$. This indicates that after performing $$$OR$$$, the answer will be at least $$$mx$$$ and it can be more than $$$mx$$$ if other $$$mn$$$ has some on bits which are off in $$$mx$$$. In the second loop, we are just turning on those off bits of $$$mx$$$ to find the final result.
UPD: This solution doesn't provide correct output for all cases.
Yes thank you that's exactly what was going on in my head but I wasn't able to put it into words
Submissions with this solution is getting uphacked.
Can you give me an example test case?
here
Weak pretests! and testers asking for upvotes
me
To not keep you waiting, the ratings updated preliminarily. In a few hours, I will remove cheaters and update the ratings again!
My brute force solution passed Div1B.
Can anyone hack it? :)
this
Hi everyone, I was stuck in Div2-C for more than half an hour. People say if you are stuck for more than half an hour, you shouldn't try more and just wait for the editorial. So, I gave up. Later when only 15 mins were left, the idea struck my mind. And I couldn't implement within those 15 mins. But after the contest, I submitted and it was correct. So, how should I decide whether I should give up or not?
Never give up in a live contest. Try the other problems, and keep thinking until there's no time left.
It's so unfair when pretests are too weak, especially in problem C, D. Luckily, I've passed the system test, still, I would say it's not fair.
code
Why this solution of O(n*m) for problem C is giving TLE on test 10
You initialize ll a[n], b[n], maybe it should be b[m]?
Thanks, Yeah it worked !!
It should have given Runtime Error?
Weak pretests in D. It hurts.
Oh... I get it. They didn't write "we tried to make strong pretests" so they aren't.
Can anyone tell why my code for div2 C is giving TLE. code
It should be b[m] not b[n]
thank you
During the contest in Div.2 C, the verdict was pretests passed and after system testing, it became wrong answer on test case 9. Can someone explain me why?
During the contest your submission is only tested on few tests. The rest of the tests are processed after the contest.
thanx bro!
Failed system test in problem C. And finding out that it can be solved using brute force. Just disastrous.
http://codeforces.me/contest/1394/submission/89728986 A hack for Div1 B
I enumerated all incoming-edge pairs of every vertex in this solution, which has a complexity of O(n^2) in the worst case.
However, it got Accepted when I submitted after contest :(
By the way, I improved my solution and resubmit for 3 times during the contest in order to fix the bug. And, my new code Failed System Test because of another mistake TAT
Wish for stronger system tests :(
Contest was amazing but Testers ruined the contest
Ok, I have been trying to debug my code for Div1B for more than 2 hrs, but I am still not able to find why this is failing. Its just a simple brute force. Can anyone point out where I am going wrong?
Your solution undercounted because "cnt == n — 1" ... a valid solution can result in more than one cycles being formed.
Here is a simple test case to illustrate:
4 8 9 1 2 1 2 1 3 3 4 2 4 3 4 1 3 10 3 1 30 2 4 20 4 2 40
The answer is 362880 (equals to 9! which means any valid combination works). Your code ouputs zero.
Ok, I got it. I assumed it will always be one big cycle. My bad, I was mislead by the test cases. Thanks a lot.
https://codeforces.me/contest/1395/submission/89733714
Can anyone help me out with Task A div 2
Removing the min from r,b,g is not always correct. If you have more than 1 odd number you only need to remove 1 from r,b,g and add it to w. And if you still have more than 1 odd number the answer is no.
Got it. Thanks a alot.
What is test case 16 of div2 D can anyone help:(
System tests ate my rating!
int main() { ios_base :: sync_with_stdio(false); cin.tie(NULL); auto startTime = curTime();
int n,m; cin>>n>>m; vi a(n); vi b(m); rep(i,n) cin>>a[i]; rep(i,m) cin>>b[i]; ll mx=0; rep(i,n) { ll mn = inf; rep(j,m) { mn = min(mn, a[i]&b[j]); } mx = max(mx,mn); } ll ans=0; rep(i,n) { ll mn = inf; rep(j,m) { mn = min(mn, mx|(a[i]&b[j])); } mx=mn; ans = max(ans,mn); } cout<<ans<<endl;Can anyone explain the logic of this solution? I saw same thing in many accepted solutions.
In the first loop, $$$mn$$$ is the minimum $$$a_i$$$ & $$$b_j$$$ for each $$$i$$$ and $$$mx$$$ is the maximum $$$mn$$$. This indicates that after performing $$$OR$$$, the answer will be at least $$$mx$$$ and it can be more than $$$mx$$$ if other $$$mn$$$ has some on bits which are off in $$$mx$$$. In the second loop, we are just turning on those off bits of $$$mx$$$ to find the final result.
UPD: This solution doesn't provide correct output for all cases.
How can you say that answer will be atleast mx?
Because $$$OR$$$ of $$$x$$$ and $$$y$$$ is never less than $$$x$$$ or $$$y$$$. That's why after performing $$$OR$$$, the answer will be at least $$$mx$$$.
Submissions with this solution is getting uphacked.
This solution seems to have a wrong output on Test 32
Which test case?
On Test 32, which inputs : 54 12 435 115 422 469 19 176 292 395 273 217 160 343 277 500 124 360 315 59 2 477 465 138 175 392 146 438 4 96 231 182 257 183 51 117 430 125 167 107 238 66 236 96 319 62 365 186 83 414 166 198 224 129 60 195 327 62 310 278 300 228 295 113 502 320 463 223
we should output 63, but this code runs with an answer 62 .
Problems from this round are really nice ones! Maybe the pretests could have been stronger. My friend's code for problem Div1C is $$$O(n^2*log(n))$$$ and it passed pretests. I'm not sure if it is a good idea to use such weak pretests for CF rounds. Besides that, this was an interesting round. I enjoyed thinking Div1B. Thank you for your effort! :)
In 1394C - Boboniu and String, there was no test with a string longer than $$$200\,000$$$ (while the limit was $$$500\,000$$$). People could use too small arrays or iterate/binarySearch in much smaller range. Come on, doesn't Polygon warn you automatically if you don't hit some min/max value?
My sad story is that I noticed my own mistake during a contest, resubmitted and thus dropped 5th -> 20th place. My first code gets AC though, and only now I uphacked it with one long string.
Rip :(
I develop app , CP helps me build logic and thinking capability.
Thanks for the contest! But a little overkill B :( Many people complain about pretests, but I think pretest were actually good (upd: good means not strong, good — finally allows to hack).
Dude, are you nuts ? How can you call the pretests strong ? Every other solution is failing in system testing. Many solutions even failed A and B in Div 2
I know it is hard skill to read, but try it again. I said "good", I did not say "strong". Also "pretests" is not the same as "tests". And good, because hacking this round was not useless feature again. But I was talking about Div 1. Div 2 A or B probably should have all tests in pretests.
Is the guy Binod?
Any reason why the multiple testcases within a single test format was excluded today? Believe recent rounds had fairly less systest fails cause of that. Might be difficult / annoying for both parties but is still the better & safer choice imo
Yes and also when I am going through the test cases it seems that the pretests were just randomly made safeguarding the actual main tests.
How do you know which of those tests were pretests ?
Just before system tests try using side bar filters present try looking for maximum test on which most of the submissions failed, that number is a good idea of the boundary condition for number of pretest. You have to do these stuffs during the contest.
Ohh..so u mean out of the N tests that are visible now (after system testing), the pretests are the first K tests ?
I initially thought that pretests are randomly shuffled in the system tests.
Yes
Your profile picture paired with your comment is perfected :o
Auto comment: topic has been updated by sshwyR (previous revision, new revision, compare).
Thank You for the Contest, I have finally become candidate master.
Congrats Bro, Your Graph is very inspiring for me
Thank You
Congratulations, Candidate Master!
Thank You
I followed you when you was specialist.
Congrats :P
In div2 A problems, I found my code is wrong.
input >
In the above test case, The answer is 'Yes' but my code prints 'No'. 89667141
(If the test case is added, Will the standings be affected?)
Standings won't be affected.
When will the editorials come out?
It is not easy to prepare a round in CF. This round is good except the test cases(not pretest but all test cases in some problems). For instance,in Div.2 C,many participant failed on test 31 which was emerged during the hacking phase. Except that,I am still greatful for the effort of writers and coordinators.
https://codeforces.me/contest/1395/submission/89766628 This is my answer to div2 E question. But there is a problem. This program cannot be guaranteed to be accepted. So here comes the question. If the program is submitted multiple times during the competition and all pass the pre-test. Which program will you use to run the system test? It is the last submitted program. Or do all programs that pass the pre-test run a system test?
The last correct submission is judged always.
OK, thanks a lot. Therefore, if a random algorithm is to be guaranteed to be accepted, it has to run multiple times. It’s just that there is a theoretically wrong probability, even if it’s small
I'm unable to understand the editorial for div2- C. Can someone explain to me clearly?
If X|Y = X then no new set bits are introduced in Y. If we have to check whether the answer is atleast X then for every i there should be a j such that (a[i]&b[j])|x = x, we iterate over x and find out which is minimum
editorial?
It has been out for some time. LINK
How to prove that the function is unimodal in Div1C(Div2F)?
Thanks
Btw did nobody notice that div1D had pretests = systests? There were some "WA on pretest 71" during the contest.
In problem 2 the vertical coordinate was X and the horizontal one was Y. Why?
In Div2 problem E, why aren't the tuple's length equal with the cycle's edge number?
Like in example one, one of the cycle is from node 1 to 2, to 4, to 3, then back to 1 with 4 edges, so the tuple should be (1,1,3,1), right? But the answer tuples are (1,1,3) and (1,2,3). Is there something that I missed about this problem's statement?
code why this code for problem C won't get TLE