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, osaaateiasavtnl., 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

it means ur solution is wrong , it passed pre test but can't past corner or tricky or large cases

This means another contest participant solved the problem, then locked in his/her solution, then went and looked at your solution, then submitted a case where the judging system and your solution came up with different answers.

Awesome Contest!! Each problem required thinking and skill set and that's what the aim of competetive programming is. Just that thepretests for B could have been a lot better.my code of problem c is misjudged and given wrong answer but on different compilers it is giving right answer. My rating has been decreased. Submission link: **https://codeforces.com/contest/1111/submission/49432400** contest link: **https://codeforces.com/contest/1111**

same problem here . on devcpp output for second test case is 8 whereas in codeforces output is 10. submission link https://codeforces.com/contest/1111/submission/49450779

what can we do now...

am i the only person whose name color didn't change after this contest..?

Should we expect an interactive problem?

No

hope it would not be like today's topcoder round ! which burnt the ***

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

Good luck

Best of luck bro <3

Hope this contest should not be unrated

Use my upvote as a "Delete Contest from Codeforces and our lives forever" button!

::::xD xD xD

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.you guys are copying minecraft

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

10writers are here.Hope to see a good contest.

Hope contest will be beautiful as it's id.

https://codeforces.com/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

Codecraft is one of the best contest created by India. It has a lot of learning stuff.

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?

HackforcesRound #537I 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

Noobs will be noobs.

(PS: I AM A NOOB. I AM NOT A MASTER)

manage*

thanks

abe tu pehle manage likhna to sikh le ghode...

u solved only 2 why ?

I guess because coding in Node.js is a real pain in the ass.

if yes , then he would not be an international master !

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.

lol : )

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 allmsmallest 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 ;)gawwwwwd level _/_

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

delwithin range [1,min(n- 1,m)] (inclusive), withdelbeing the number of deleted elements.Let's denote

sufsum(i) as the sum ofilargest elements of the array.For each iteration, since we have used

deloperations as deletion, the number of increasing operation should bemin(m-del, (n-del)·k) -- we'll denote this value asinc.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.

just use long long array to store sufsum

anyway thanks for the explaination

LOL I was not aware of the fact that u cannot resubmit a hacked problem :(

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

x^{φ(p) - 1}always has remainder 1 when dividing byp, provided thatxandpare coprime. φ(p) is Euler's totient ofp.In this case, since 10

^{9}+ 7 is a prime number, then φ(10^{9}+ 7) = 10^{9}+ 6, thus instead of dividingb!, you will multiply , taken modulo of 10^{9}+ 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(

n^{2}) algo get TL on max test (n = 10^{5}, m = 10^{7},a_{i}!=a_{j}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.com/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

mine was after that contest : Prakash_Gatiyala 2019-02-03 20:36:37MSK(21:06:37IST) Announcement Hack announcement ***** Unfortunately, your solution on B has been hacked :(

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 .....

another one : Hack test case for B: 5 4 3 2 2 3 3 4

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

more than that , i couldn't get any wind until last what d is asking moved to e . get blown . lol : ) btw a good round with huge difficulty gap , even c was not easy

It's in the statement:

Yes, Y is

nota 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 2

^{30}? 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·log^{3}(2^{n})) which is like 2.7e9.Solve is called (

k·log(2^{n})) times, and each call (discounting recursion) takesO(log^{2}(2^{n}))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

everytype 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 * logKbut it got TLE on pretest 8. https://codeforces.com/contest/1111/submission/49423653You can see my opinion below // In fact i don't know whether it's true or not // <sorry i'm bad at E>

Yeah mine soln with the same complexity gets TLE at pretest 8 , very strict TL for JAVA :( My soln : https://codeforces.com/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

gofunction, 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.Still it gets TLE'D :( https://codeforces.com/contest/1111/submission/49440697... I just did the memoization and btw the operations in JAVA's HashMap is O(1) amortized so no extra log factor .

someone can tell me : Is there any wrong in my way in C please : Firstly i add all the index of a1..ak and sort . After that i check whether i can put a 2^k number among them and sort again Now i can cut the array into many segments and count Number_use A = the number of bit 1 in subtract distance of 2 continuos index // and the rest is use B independently // Tk you so much

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

freq_{c}is the frequency of charactercin strings,Kis 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 . Ifs_{x}≠s_{y}, there is an additional condition thats_{x}ands_{y}must belong to the same subset.Let

fl[c][j] the number of ways to split the set of characters from 1 tocinto two subset, such that sum of frequency of the first subset isj.If

s_{x}=s_{y}, thenKis simplyfl[c][n/ 2].Consider the case when

s_{x}≠s_{y}. We need to precalculatecnt[c_{1}][c_{2}] — the number of way to split the set of characters into two subset whenc_{1}andc_{2}must come together — thenKis equal tocnt[s_{x}][s_{y}].Precalculating

cntcan be done naively inO(n*ALPHA^{3}) (ALPHA= 52 in this problem), which is not fast enough.A faster way to do this is as following:

Iterate over all

c_{2}. For eachc_{2}, calculatefr[c][j] — the number of way to split the set of characters fromctoALPHAexcludingc_{2}into two part, such that sum of frequency of first part isj. Then, one can iterate over allc_{1}, combinefl[c_{1}- 1] andfr[c_{1}+ 1] to getcnt[c_{1}][c_{2}]. Complexity isO(n*ALPHA^{2}).Thank you, loved the trick.

I had a different

O(n* α^{2}) 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 setDP_{new}[j-a] =DP_{old}[j] -DP_{new}[j+a], looping down fromj=n, whereDP[j] is the number of ways to get a difference ofjto the sizes of the sides. We can undo two characters inO(n), so this algorithm is alsoO(n* α^{2}).Code: 49428256. Here

DP[n] is zero difference, and less thannis negative difference.Feels like i accidentally enter to Codechef

haha. how u solved c ?

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.com/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()`

or`range()`

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?

yup! ! , this yr , more difficult . one ques. why u didn't solved B ?

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

allcharacters ofeach typelay 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

Pretests were not properly framed.unrated.I find

allpointsvalid.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.

that's why don't discuss b4 st

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.

Well I participated in contest and solved first 3 successfully. I just didn't want my main ID to have negative contribution coz I knew people don't wanna hear truth.

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.

hahhahaha . i got wa at tc 105+

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

those who passed examples only will get AC

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!

these are thhe signs of unexperienced setters. i got wa at 108th tc of b . just bcoz pt was soooooo weak , if b would have been c , people would have been more cautious

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.

its weird that no body is discussing problem E . only with some D . this condition i see in div1 rounds where only legends pass from c to d to e .

what is happening to div2 rounds .\ MikeMirzayanov

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 :)

Make it unrated

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.

till now only 4 official(div2) submissions accepted for problem 4 and 0 for e . i think problems got swapped with div1d and e /

can anyone tell me why this is giving TLE ?? https://codeforces.com/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.

my code of problem c is misjudged and given wrong answer but on different compilers it is giving right answer. My rating has been decreased. Submission link: **https://codeforces.com/contest/1111/submission/49432400** contest link: **https://codeforces.com/contest/1111** @nitesh_gupta

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

eachelement k times.Sadly I got it otherwise during the contest :c