Hello Everyone!
I would like to invite you to CodeCraft-19, which is a part of Felicity, the annual techno-cultural fest of IIIT Hyderabad. It will take place on Feb/03/2019 18:35 (Moscow time).
The round will be rated for Div 2 participants (whose ratings are lower than 2100). As usual, participants from the first division can participate in the contest out of the competition. You will be given 5 problems, all of which are based on the theme of our fest, "Superheroes".
I would like to thank the CodeCraft team -- nir123, additya1998, devanshg27, gaurav172, night_fury208, shaanknight, psaini, sharmaritvik60, swetanjal and vivace_jr -- for helping in preparation of the problemset. I would also like to thank KAN and 300iq for round coordination, ----------, Aleks5d and mohammedehab2002 for testing the round and MikeMirzayanov for the great Codeforces and Polygon platforms.
Following the convention, score distribution will be announced shortly before the contest.
Good luck to all the participants!
UPD1: 500-1000-1500-2000-2500
UPD2 Editorial is published! We sincerely apologize for the weak pretests in B and for the difficulty gap from C to D which turned out to be harder than what we expected.
UPD3 Congratulations to the winners!
Auto comment: topic has been updated by nitesh_gupta (previous revision, new revision, compare).
poor pretest
I Didnt understand what is meant by when it says your solution is hacked
Should we expect an interactive problem?
No
Ohh thanks
Just curious. Why not div1 this time?
There's less number of people who want to set problems for Div. 1 because they're harder.
Wishing everyone high ratings.
I like when people put up handle like these lol.
Some previous editions of CodeCraft:
CodeCraft-18
CodeCraft-17
CodeCraft-16
CodeCraft-15
Indian round?
TShirt? :/
Prizes are not allowed for Div2 contests as it may lead to fake accounts by Div1 users.
aryanc403 surely meant overall winners rather than Div 2 winners. There is no incentive for fake accounts in that case.
Good luck for all participants.
Maybe this contest is lucky for me to become pupil
Hope this contest should not be unrated
Use my upvote as a "Delete Contest from Codeforces and our lives forever" button!
Expect numerous hacks like last time.
I feel sorry for all the hacks, but I at least I can say I warned you beforehand.
GL & HR Everyone.
.
Mancity & arsenal or codeforces :'(
Codeforces round of course!
I want to be blue but it will never happen.
Same as me
good luck!
i hate cyan already
Were there alot of hacks last time?-
Hope, problems have harder pretests to prevent so many hacks like the previous year!
The pretests are quite weak, and it seems there are more hacks than last year :C
Not only hacks but also system testing failure!
How long will contest held??? 2 hour or 2.5 hour ?? not mentioned in post.
2 hours
10 writers are here.
Hope to see a good contest.
Hope contest will be beautiful as it's id.
https://codeforces.me/contests/1111
contest 1111. It's really a nice number. woof woof woof.
比赛1111 这是一个非常美妙的数字 汪汪汪
I hoped that there would be a real Lunor Year contest tomorrow... 我曾经希望明天会有一场除夕赛。。。
"Jarvis has to tell Iron Man the number of distinct colony arrangements he can create from the original one using his powers such that all villains having the same type as those originally living in x-th hole or y-th hole live in the same half and the Iron Man can destroy that colony arrangement."
This run on sentence is murdering my brain.
After doing ABC and not finding out the reason why D is failing, all I can do is my part
I vote for unrated
chup chutiye..
I have no idea what you try express with those words. Translate please
It's so nice to watch the solution go through 30 tests in a row in C
shit contest kys.
make it unrated or die
worst constest,shit pretest ,more than half of B'solutions in my room fail with the same testcase. pd: my code alse fail in that testcase :(
Do you happen to know what that test case is?
Mine failed because I didn't check for m < n — 1.
I used that testCase
4 2 1
1 1 1 100I hacked 5-6 solutions with :-
4 2 2
1 3 3 3
I used 3 1 1 4 3 3 to hack 3 solutions in my room
Is it HackForces?
Hackforces Round #537
I always wonder why 'Codecraft' is challenging every year. Each year you get to see challenging problems on some particular theme. I am sure we'll find some nice stuff to learn from its editorial too.
lol how did you manage to fail A and B? noobs
Maybe python could help here!
manage*
thanks
u solved only 2 why ?
I guess because coding in Node.js is a real pain in the ass.
check your B)))
as it was said before, node.js is like a needle in the ass
With wrong statements (On B)
HackCraft-19
What is Test 6 of Prob C?
It's a test where probably there are one or more avengers on the same position. Think if your code works when you have duplicates.
More than one stay in one place :) Lost 50 mins for this.
can you please share your approach?
Reccurence.
will it not time out?
I think not time out...
if possible can you please suggest a counter case for this
Reccurence should work if you cut it immiedietly after current interval contains 0 avengers(result is A then).
This test helped me
This round was so weird...
so weak tests...
Really liked problem C :) But it was kinda a typing contest though :(
Classic shitty round
how to solve Problem D
Personally I dislike this contest.
Codecraft in two most recent years have been plagued with weak pretests, and to top it all off, this year has difficulty spike and incomprehensible + mistakeful statements along.
Well, honestly saying, the reduce from a Div.1+Div.2 combined to a Div.2-only is enough foreseeing a major drawback. Hope to see better attempt next year. ;)
Why didn't it (49418380) get RE on test a ab ?
Problem D statement clearly looks like it is asking about something, while it seems that it is actually asking about an entirely different thing.
I got a solution after some time, then the clarification destroyed it. What was in the clarification wasn't clear in the statement.
I have just noticed now that there was a clarification. LOL it added an entirely new condition that wasn't mentioned in the statement.
5 4 2 4 7 9 11 100 problem B test XD sadly mine also failed...
What is it supposed to be? I get 40
correct!!!
Pretests were a disaster! Many sentences in different questions were unclear and tricky, especially B.
What were the hacks for A and B
For A I mostly used:
For B I only used:
Thanks — what were you looking for in the code for those hacks to work?
For A it was easy. I just looked at code, quickly understood that someone is only counting occurrences of vowels and consonants and compare them, looked at the statement to make sure that it is wrong and created this simple counter example.
For B the codes were generally harder to understand but when I found a 5 lines of code solution I realized that many people assumed that its optimally to firstly remove as many smallest elements as possible and then increase the biggest guy. This test is a counter example.
Some tests I used for B:
What did you notice in the solutions that made you know the hack would work?
The first test I used was to counter a greedy solution, in which if m ≤ n - 1, such solution would delete all m smallest solutions.
There might be an optimization, in which the deletion will stop when reaching all remaining elements have the same value equal to
max
. That's when my second test came to the play. ;)is the answer for the second one 8.714?
Yes, the optimal will be increase any element
9
once ;)what is the ans for given 2 cases?
For the first test:
9.5000000
For the second test:
8.7142857
what is the solution for B?
I approached it with a bruteforce solution ;)
Obviously, if we're to remove an element, that must be the currently lowest element. We'll take that into regards, and sort the array in an ascending order.
Now, iterate through all del within range [1, min(n - 1, m)] (inclusive), with del being the number of deleted elements.
Let's denote sufsum(i) as the sum of i largest elements of the array.
For each iteration, since we have used del operations as deletion, the number of increasing operation should be min(m - del, (n - del)·k) -- we'll denote this value as inc.
Thus, the answer for each iteration would be .
We'll take the maximum value throughout all iterations as our final answer. ;)
i have done the same.... still got WA at test case 15 49417970
m-i might become negative.
anyway thanks for the explaination
WORST CONTEST I HAVE EVER SEEN ON CF.
HACKERS!!! HACKERS!!! EVERYWHERE xD
How do I calculate ((a!)/(b!)) mod p, where p is prime?
using the fermat's little theorem
a == a ^ (p — 1) (mod p)
therefore, we can find the modular inverse by
a ^ -1 == a ^ (p — 2) (mod p)
and the result is
a! / b! == a! * b! ^ (p — 2) (mod p)
Sorry for poor presentation, I just dun wanna type any latex
https://www.quora.com/How-do-I-evaluate-A-B-mod-1000000007
You can use the Fermat-Euler theorem, which shows that xφ(p) - 1 always has remainder 1 when dividing by p, provided that x and p are coprime. φ(p) is Euler's totient of p.
In this case, since 109 + 7 is a prime number, then φ(109 + 7) = 109 + 6, thus instead of dividing b!, you will multiply , taken modulo of 109 + 7. ;)
This is the contest with the weakest pretest I've ever participate
mistaken simulation task as greedy / DP for the last two rounds, solves it later than everyone else.
Life is sad
difficulty gap again too high ! even many yellow coder solved only 3 and that also in division 2 . division 2 is smoothly becoming div1 .
Border Test Cases for B on which many solutions will fail ! 5 2 2 7 7 7 7 7
5 2 6 7 7 7 7 7
Hack for B 5 4 3 2 2 3 3 4 Answer is 3.666666
can anyone explain problem B
As always, everyone solved the first n problems, and some unicorns solved the rest.
haha! unicorns !
Is the intended complexity of D is O((52)^2*n) ???
So you tell me solving ABC gives you 2nd place, while solving ABCE gives you only 14th place? Not cool.
Edit: A(failed B)CE and 37th place
Why didn't this(49426136) O(n2) algo get TL on max test (n = 105, m = 107, ai != aj for all i != j)?
Problem D: can anybody tell me why my code is WA?
I used fast binomial coefficient calculation using precalculating factorials ans factorial inverse.
half = n/2, nx = number of char x in string s.
a, b, c, d ... is not targeted chars.
I just calculated 2 * C(h, nx) * C(h-nx, ny) * C(n-nx-ny, a) * C(n-nx-ny-a, b) ...
And I also treated s[x] == s[y]
PLEASE HELP ME!
http://codeforces.me/contest/1111/submission/49432460
aabbcc 1 1 2
answer is 0
OOPS.. I think I misread problem. Thank you!
Hack for B
5 4 3
2 2 3 3 4
Ans is 3.666666
Hackforces again.
When your solution was hacked in the last minutes
Nvm, it happens every time on codeforces
When one remember that all what he did in the life is keeping on brushing his teeth.
I hacked someone intentionally in the last minute
How to solve problem C?
another hack round :(
Hack Case for B:
3 3 1
9 9 10
Expected ans: 9.6666
In my case, i got w/a this testcase.. T^T
(Good) But Many output of Pretest Passed Code will be 9.5 .....
Added setters to black list.
Y is not a vowel? And the pretests didn't test that??
I liked D/E, but their difficulty was more like div1 D/E :P
It's in the statement:
Yes, Y is not a vowel, at least in English. It can sometimes be considered a vowel but it is primarily a consonant. Sure, normally on codeforces 'y' seems to normally be classified as a vowel but it's slightly more correct to have it as just a,e,i,o,u if the statement is in English.
"where each position can be occupied by many avengers" i m not able to submit c because i didn't observe it
In C, I was using feynwick tree to find number less than x(not binary search). Got TL on test case 5. TL seems be too strict.
So you had a Fenwick tree of size 230? What's odd about that TLEing?
I stored map. Link. It was O(log^2(n)).
Your solution looks like it would be O(k·log3(2n)) which is like 2.7e9.
Solve is called (k·log(2n)) times, and each call (discounting recursion) takes O(log2(2n))
Your method must be wrong, mine passes in 78ms.
The most imbalanced round I have ever seen, literally everything depends on speed and hacks.
1-based indexing caused 2 of my solutions to fail... (for I use C/++)
I found statement for D ambiguous.
"Iron Man can destroy a colony only if the colony arrangement is such that all villains of a certain type either live in the first half of the colony or in the second half of the colony."
I think something like this would've been better:
"... only if for every different type of villain, all of its occurrences are either in the first half or the second half of the arrangement."
Not specifying that the property should hold for every type of villain was misleading. I personally misunderstood the problem it until the clarification came up.
I second this. The entire contest, I understood the condition as "there exists a certain type, such that all villains of this type either live in the first half or in the second". Unfortunately, the samples did not differentiate between the two cases.
When the clarification came up, I notified the authors of the possible ambiguity in the original statement. But I was not sure if it was just me or if the statement was indeed unclear. Glad to hear it wasn't just me :)
Thanks, that makes the problem more clear :)
Really unfair for a guy(say x) who has solved 3 faster than another guy(say y). X and Y lock their solutions but X finds no hack in his room and Y finds a lot of hacks in his room. Its unfair on the side of X though. Pretests should be strong and Questions should be balanced a lot more than what it was today.
You are right in that pretests should've been stronger. I don't agree with the rest of your comment. Rooms are selected randomly, so everyone has the same chance of being the lucky Y guy. That's just how CF rounds work, sometimes luck is an important factor as well.
I would prefer getting ratings by solving rather than luck :)
It sucks when someone beats you out of luck, I've been there too. But that doesn't mean it's unfair. He/she didn't break any rules, so we should accept it, even if it's hard. Moreover, if that lucky guy had been you, you still would've won and gotten your rating.
That's how mafia works
Is question c can be done with the help of segment tree and dp with memo???
just use simple D&C
No! even if you handle a[i] <= 1e9 u will get TLE.
I believe the complexity of my submission (in Java) for C is N * K * logK but it got TLE on pretest 8. https://codeforces.me/contest/1111/submission/49423653
Yeah mine soln with the same complexity gets TLE at pretest 8 , very strict TL for JAVA :( My soln : https://codeforces.me/contest/1111/submission/49427480
Inside the recursion, the first thing you should check is if there is any index between l and r. If not then return A
Did you checked my code properly?, I did return 'a' for this condition.
aman110599 TLE must be because while executing the go function, you are checking for memoization but if you think properly then same states won't be reached again. So there is no need of memoization and you can remove that line hence remove the extra log factor.
That jump of difficulty from C to D... It's like playing Dark Soul after learning how to use the computer.
It’s like asking barbarian to play dark souls actually
what was your solution for D ?
The answer for each query is , where freqc is the frequency of character c in string s, K is the number of way to split the set of 52 characters into two subset, such that sum frequency of characters in each subset is equal to . If sx ≠ sy, there is an additional condition that sx and sy must belong to the same subset.
Let fl[c][j] the number of ways to split the set of characters from 1 to c into two subset, such that sum of frequency of the first subset is j.
If sx = sy, then K is simply fl[c][n / 2].
Consider the case when sx ≠ sy. We need to precalculate cnt[c1][c2] — the number of way to split the set of characters into two subset when c1 and c2 must come together — then K is equal to cnt[sx][sy].
Precalculating cnt can be done naively in O(n * ALPHA3) (ALPHA = 52 in this problem), which is not fast enough.
A faster way to do this is as following:
Iterate over all c2. For each c2, calculate fr[c][j] — the number of way to split the set of characters from c to ALPHA excluding c2 into two part, such that sum of frequency of first part is j. Then, one can iterate over all c1, combine fl[c1 - 1] and fr[c1 + 1] to get cnt[c1][c2]. Complexity is O(n * ALPHA2).
Thank you, loved the trick.
I had a different O(n * α2) solution, but it's more complicated than yours.
Due to the way the DP is calculated, we can undo any character from the DP. So just calculate the DP in O(n * α), then for all pairs of characters, undo those characters, and check from the resulting DP-array the number of ways to split such that the difference is equal to the sum of their frequencies.
To undo a character with frequency a, you set DPnew[j - a] = DPold[j] - DPnew[j + a], looping down from j = n, where DP[j] is the number of ways to get a difference of j to the sizes of the sides. We can undo two characters in O(n), so this algorithm is also O(n * α2).
Code: 49428256. Here DP[n] is zero difference, and less than n is negative difference.
Feels like i accidentally enter to Codechef
Failed to get up on time and it seems a typing&hack contest... Hope not to FST!!
In problem C, manual binary search code gave TLE on pretests while use of upper_bound and lower_bound passed the pretests! Shouldn't happen like that , very sad to see this. :(
But that is not strange.
Doesn't your comment just reduce to "I wrote binary search incorrectly"?
how u solved c ? this was very weired conterst , ur a and b failed st , bcoz of so weak pretest .
Well, my A and B in particular failed because of really dumb bugs that I should have caught myself. I won't blame the pretests for my poor implementation mistakes.
As for C, I did DFS on trie, which afterward I realized was overkill lol.
many solved with recursive approach using lower & upper bound .
Yeah, I realized that was possible after the contest.
divide and conquer + purning
mujhe mat sikha bc
LOL. But he is right.Your bad
My Submission Link
Please say what could i have done more
I just added
else return temp
to your code. 49440423Thanks very much
I solved C using dynamic segment tree. Unfortunately couldn't code before time, as my B got hacked and spent lot of time finding the bug.
Have to wait till system testing to submit.
Can someone tell if dynamic segment tree is fine for AC?
Yes, it is. It's unnecessary though. Since there are no updates, you could use a map for storing prefix sums. It's a lot easier to implement and has the same complexity.
Using map for storing prefix sums was too slow for me. https://codeforces.me/contest/1111/submission/49442474 Any optimisation I couldve done?
I think it should be fine. But you can notice that you don't need that segment tree because there's no update. You can just do a binary search to find how many elements are there in this segment.
what is dynamic segment tree?
Usually in a segment tree, we initialize the whole tree with nearly 4*mx nodes and operate on it. Where mx is the maximum value of a[i].
But what if a[i] takes values (1 <= a[i] <= 10^9), but there are only 10^5 elements in array a[].
That time we can use dynamic segment tree.
We only create a node of the tree if its going to be affected by our
update()
orrange()
query. Else we don't require it. Thus making it "dynamic".You can check my submission for problem C in today's contest for better idea.
thanks for explanation, i will try to understand
Problem C just minced the JAVA users , as who keeps the TL for a problem with recursion to 1 secs ?
Hosted on wrong platform typical codechef contest
Has anybody solved C with dynamic segment tree? My submissions got MLE #8
My dynamic segment tree got AC.
I think you are doing too many unnecessary allocation of memory ( i.e ptr = new node(); ).
Please check my simple implementation.
Thank you for your code. Here is my AC submission 49472118. I got rid from the
rec
function and did all the calculations in the segment tree.Hmmmmmmmmm
For like 90% of people y is a vowel and you don't even put a pretest that should get it wrong? Aww come on
For A, vowels were given in the statement.
Like 100% of people know how to read a statement.
Is Codecraft always like that?
I doubt there were >EPS different pretests in B
POOR CONTEST .....DO YOU WANT TO JUDGE ON THE BASIS OF HACKS ....EXPECTED MUCH MORE FROM IIIT — H
Any verdict, system tests.
I doubt if mining bitcoins is a greater waste of electricity than running today's pretests on Codeforces machines.
haha
Make this contest unrated, The codeforces site didnt respond for last 4-5 minutes. It was showing Error 403 Forbidden. Was not able to make submission.
In problem D, Ironman can destroy the colony if and only if first half or second half is "full of certain types"?
for example "aaabcd" and 'a' is a certain type, we can destroy it. but "aabcde" ans 'a' is a certain type, can we destroy it?
I can't understand problem...
For the first question: No.
For the aabcde:
Yes we can, because all characters of each type lay in one of the two halves:
Type 'a': each character that equals 'a' is in the first half.
Type 'b': each character that equals 'b' is in the first half.
And so on..
Thanks for the explanation!
Statement ambiguity for D is discussed in this thread
I find all points valid.
Another hack for B
3 100 1
1 2 22
answer=12 there is test 15
Pls, someone teach the coordinators to say "No" to such tasks and tests. What were the testers doing during preparing the round? Make this shame unrated, and let no one ever remember about it
Yeah exactly, I would like to forget this contest.
Fully agree with you
As system test proceeds, the contest is turning out to be a speed test for problem-A. This is unfair.
Please make it unrated.
100+ test case in b , thats weired with so weak pt .
During system testing, they add all the hacks that other users submitted to the tests.
Yeah so basically,
For a not-pro coder like me:
1) Solve A.
2) See B, think too much.
3) After a while, see number of submissions increasing fast.
4) Guess a solution and get prestests passed.
5) See C, think too much...
6) Unfortunately your B has got hacked :(
7) Try fixing B. Get it fixed somehow. But couldn't solve C because of time.
8) Fail main-tests for B.
9) Rank list = Speed test for A :)
Sometimes 6) is omitted.
I meant in this contest particularly. Not in general :)
I meant that, too. -> (In this contest, somtimes~)
Its turning out to be efficiency test for B. Obviously the ones who solved B thought better than others. And that's what should affect rating coz rating is all about how well you can solve a problem.
I think its easy to talk without participating in the round.
In D , the lenght of S should be <=10^4 it will be more fair
Now I know that in iiit-h, h stands for hacks!
Anybody has an idea for test 115 in problem B?
That was my testcase for hacking. I got 5 hacks in my room with it.
I propose that anybody who passes the first 100 test cases on problem B is automatically counted as Accepted.
But why? Their solution is wrong if it doesn't pass all tests.
Sure, but if we get rid of all the tests past 100, then the solution will pass all tests and be correct.
Yeah, and if we have no tests for any question, then everyone can have AC.
So fair! What if you got the wrong answer on test case 90? :)))
Ok, let's make it so that anybody who passes the first 89 is accepted then
I like you got wrong on 111. Let's stop thinking about our ratings and getting accepted while our solutions are wrong. The explanations and pretests were really weak, don't kill the facts just for your sake!
Authors, stop doing contest please!
Shit just got real! IIIT-H Thanks for hackforces!
The Rating of Problem B will go up...
Yyyyeeeaahhh boooyyy!!!
4.3k for A -> 550 for B. The biggest jump I remember.)
Thanks for weak pretests!
why so many WA on B ?
There should be much precision (I don't mean 0.000001 precision) in a solution of this problem, beside that, the weak pretests will result in these many WAs, because one will not pay attention that his solution has weak code when he gets Pretest Passed.
I don't think that precision had anything to do with it. But most people submitted an incorrect greedy solution.
For Problem B, the codes that are treated as Wrong Answer on test cases 109, 110, 111, or 115 appear frequently. Did the Problem Setter intentionally not include these test cases into the Pretest?
Maybe it's just people's hacks being added before system testing?
Then the authors didn't have tests for those edge cases in system tests themselves? If so, does that mean they also didn't account for these edge cases (and maybe thought the problem was supposed to be easier).
Seemingly. In fact, the situation where hacks actually filling into the gaps was known before.
Probably. Testcase 115 was my testcase for hacking.
This ought to be unrated.
That round deserves and oughts to be unrated, or semi-rated, cause it's become truly unfair
Hello everyone, you know, there is such wisdom that the blue ones do not make up the rounds. Today I made sure that if they make up it is some kind of shit but a round is slop-top (sorry for my English)
but problem setters were purple too ! what about them
The very fact that the blues were involved in it is already shit, although people who are divas 2 in my opinion have no right to compose a round
That feeling, when you know your solution will fail main-tests. But still is in system testing queue :)
The number of Accepted code for Problem B is far less than that of Problem C... And is about 13% of that of Problem A now. Isn't it abnormal?
Yes.
can anyone tell me why this is giving TLE ?? https://codeforces.me/contest/1111/submission/49428534
Your algorithm complexity is
O(N * 2^N)
In the case : num == 0, you don't need to recurse because the answer will be automatically A This will stop recursion in all segments containing no elements :)
oh thanks !!
I wish this contest would have been swapped with the previous one.
- The previous contest deserved to be on the day before chinese lunar new year and also being a very good round.
- And this contest deserves to have a long queue, leading to an unrated round.
In the previous round, the cheliks crap off the queue
Was testcase 15(Problem B) intentionally not kept in pretest?
For which problem?
people who failed in B and blaming for weak pretests, better next time write proper code and don't complain. LOL
hmmm people who didn't succeed just need to succeed next time.....
UNRATED!
System test on B was hell of a massacre.
Bad pretests destroyed good problems ...
The problem difficulty and acceptance rate are very similar to the contests that happen on codechef. This has been one of the major issues for codechef and now codeforces too. Hope they get done with it soon.
why the answer of testcase # 115 is 5.500000 ?
It is 5.66666666666666666652.
Because you can increment each element k times.
Sadly I got it otherwise during the contest :c
maybe you didn't consider the fact limit of k was for each superhero, for # 115 best is 5 6 6
My first solution C with "compilation error" in contest https://codeforces.me/contest/1111/submission/49430877. The same solution C with "accepted" after contest https://codeforces.me/contest/1111/submission/49435696. Why does the same code have different result????
у CodeCraft раунды всегда ********* ** * * ***** ** * ***
Lmao this surely deserves an upvote
What is this context? I feel that I sort-of understand this. But is there a link to such a previous contest?
https://codeforces.me/blog/entry/61383
Jokes aside, this contest resulted 2342 successful hacks in total, which is less than Codecraft-18 last year (3442) but more than Codecraft-17 (2334). Also, it did not broke GreenGrape's record of 3773 hacks. It's in the top 10 most hacked contest tho.
I used this javascript in case you are interested.
the man who posted test case 110 of B is very genius
Contest should be rated.
You are Useless
The framing of questions was unclear. That is the minimum standard a person expects from a CP question. On top of that weak test cases.
What is the use of giving pretests if strong cases are not included in it?
The worst contest in my life.
mine too! :(
This is what you call 1080P shit
When your B's solution get hacked while hacking others A.......SAD LIFE!
Is it ok that solution got CE on system tests but passed pre tests?
Verdict says
Can't compile file: Compiled file is not a valid executable. Probably, the source tried to use too large static array(s).
I don't know how works GCC array extension. Maybe in this particular test just too much memory trying to be allocated...
P.S. Read that in GCC
double a[n + 1]
allocates arraya
size ofn + 1
on stack. Pretty sure it's should be SF on bign
and UB on middlen
. But why on Earth it is CE don't know.P.S. Now I'm think that I'm wrong. 49433475 and 49438205. Absoulete same code, same compiler, no any static array but first submission gives CE second AC.
u have renamed CODEFORCES into HACKFORCES !
Besides the pretests, system tests for B are also weak. see this accepted submission.
And hack case
2 100000 3
1 1
probably need more corner cases
Agree. I see a lot of submissions failed on hack tests (test >= #109) during system test. They would have been accepted without the hacks. Corner cases should have been considered thoroughly rather than adding bunch of random tests.
I think it’s a bit rudely to downvote this contest.I mean,problemsetter spent hours to make this contest,and people downvote it
I never expected such a contest from IIIT Hyderabad :(
It feels weird to be bashing on an Indian contest without Ashishgup.
Why were almost all tests for B and C(maybe other problems too, didn't check) literally random? For example, there is no max test for C, which is probably the only reason why 49441838 gets AC.
https://codeforces.me/contest/1111/submission/49427048
This submission shows compilation error. Although it runs for 39 test cases fine and shows the error at test 40. Also, I resubmitted the same solution after contest. It gets accepted. What can be the reson behind this ?
https://codeforces.me/contest/1111/submission/49433444
I maybe got same error! But, sorry, I don't know what it is and what should we do.
emm... Can anyone give a explanation about the rest examples in problem D to me? I think that the statement for D is too confusing... XD Thanks!
Problem.D means that the same kind of villains should be on the same side(left or right)
ps: they don't have to be together
for example: "abacdc" is ok!
so...
why in the first case, 2nd question's answer equals 0: because AAaa have 4 characters and they can't be gathered in the 1st or 2nd half(the side's length equals 2)!
5~8. just reverse the strings above!
thanks very much!!! :)
Have a nice Hacking last night. But Problem B failed on system test.
昨晚叉得好爽 但是B fst了
My Submission during the Contest failed with compile error, but after the contest, I send same code and accepted. It must be system error but I don't know what should I do. Does anyone know what I should do or how to contact codeforce's support?
Please make strong pretests, it will be helpful for many contestants.
Check this test case for Question 2 :- Input :- 2 5 1 1 100 Output :- 101 Correct Answer:- 100 This is output of my friends Code which passed all test cases after contest.
http://codeforces.me/contest/1111/submission/49430322 Hack me, please.
I misunderstood the statement of problem B but still passed pretests and got 4 successful hacking attempts on B >.<
Same, I bet authors didn't even have the correct solution for B
Can anyone help?... :( My code seems almost similar to the author's answer code but the memory keeps exceeding at case 6. I couldn't find what the problem is so I'm asking you guys out there for help...ㅠㅠ Thanks!
49458160
and your original code will not terminate (stack overflow)
And some more bugs of CF.
PS: Maybe CF should add foulder 'report bug'.
Why my code is compiler error in system testing, but after that I resubmit the same code, it becomes accepted again?
Why rating will be recalculated? Any particular reason?
And why this decision has made after calculatingit? Normally it is stated in announcement just after contest.
Seems like updated. And I've the same rating as before. Thanks.
Yes, me too. But what was it anyway?
No idea. There wasn't any post about that.
Some participants got unexpected wrong "compilation error" verdicts on system tests (after pretests!). I've fixed the issue.
Thank You
Thanks MikeMirzayanov for recalculating and saving the earth! :)
wait wait, there is a difference between "Rating Recalculation" and "Unrated" ..
got upset! sorry! :(
Ratings for the round 537 will be recalculated and returned back. ??? WHY
There was no reason making the contest unrated just because pretests were weak is that what you got? Even if pretests are weak its a coders duty to check his solution with his own test cases and read the statement carefully and if hacked just accept that his solution was wrong. Making the contest unrated when wished is not the right thing to do. Although i agree many people got hacked in B i myself got hacked but that's just teaches us to check our solution before submitting whatever happens. I became expert after this contest and was happy and celebrating but now this. Its just devastating. I know nothing can be done now but hope nothing happens like this in future.
it is not unrated, read the announcement, it will be recalculated
make this round unrated .
True coder is who can imagine all test cases before typing code , not making hit and trial . And problem setter can't set all test cases in pretest.
this z not reverse coding
Poor pretest ....
When your submission for task B was hacked
rated, unrated, semi-rated, ... and now we have rerated round!
Why this round problems do not have difficulty tag?