Привет, Codeforces!
Рад пригласить вас на Codeforces Round #421, который состоится 27 Июня, 2017 в 17:35 MSK.
Да, это мой первый раунд, и да, очередной раунд от фиолетового.
Хочу поблагодарить KAN за огромную помощь в подготовке, danilka.pro за помощь на начальных стадиях и тестирование, Belonogov, WHITE2302, hloya_ygrt, Perforator за тестирование задач, а также MikeMirzayanov за Codeforces и Polygon.
Как обычно, участникам обоих дивизионов предлагается 5 задач и 2 часа на их решение. Разбалловка будет объявлена ближе к раунду.
Good Luck and Have Fun!
UPD1: Разбалловка стандартная для обоих дивизионов: 500-1000-1500-2000-2500
Мы работаем над проблемой в задаче A первого дивизиона. Мы опубликуем решение позже.
Из-за проблемы с задачей A, к сожалению, раунд нерейтинговый. Подробнее.
Несмотря на все проблемы Рабор должен быть, поэтому он здесь.
offtop
Когда результаты отбора на Сазанку?
Is it rated?
This is first time that this question is logical to be asked :)
.
you too:D
We should thank KAN for his help in the last 10 rounds at least =D =D
Thank you so much :)
it's just because KAN is the coordinator of rounds.
Hope for short statements like the blog <3
Hope for short statements too. <3 Good luck for all.
Hope for short statements like the blog <3 *3
exm???
I think ImAliWithTheWAs is very lucky because his feedback still positive. I hope you good luck in contest my friends <3 .
Actually lolzano and Cydiater got downvotes because they copied his comment.
anime?
Everyone still wonders: Is it rated?
Codeforces rounds are rated by default. If a round is unrated they will write that in the announcment.
Believe me, you won't get upvotes with that way!
lord saves CF servers today, just submitted a solution and it's a 2 page long queue.
Round 420 is also "rated"...
i hope this time problems does not have any bugs good luck every one :D
**The Rain Of Downvotes :P **
Hope we can have a better exam environment and problem statement :P
When you come to increase your contribution but gets downvotes instead.
Why would you "come to increase your contribution"? Do you write comments just for meaningless internet points? Contribution should be the result of your comments, but not the reason for it.
Hope for better performance of the server..
If I were you, I would never comment lol :P
You are the only one who will comment
haha I want to get my old contribution back :P but every time I comment it gets worse lol
HelOO
yup ?
Personal talk
How do the screenshots help exactly ?
He needs screenshots for the time when he is still Nutella in case he lose Nutella this contest.
"Nutella" as in ? P.S. Why all the downvotes ? I asked a genuine question.
legendary grandmaster!!!
PS: you might understand it better if you see this.
A lot of of solutions in the status page got Denial of judgement :( !
Is the issue in the site ?
BTW, its the same solution but a lot of people try it to know why it got Denial of judgement.
The comment is hidden because of too negative feedback, click here to view it
F**k CF servers just before the contests :/
Make a wish for Akagi with 200 rating.
Well, due to the unrated, maybe I have lost more than 100 rating.
But...
Where is my Akagi?
(ノ=Д=)ノ┻━┻
RUA!
Is my English weak？i have read 20minutes still cannot understand what the problemC want to tell me。
Пиши контест говорили они, будет весело говорили они.
My story since past few contests:
1. Eagerly wait for contest to begin.
2. Contest starts. Open problems.
3. All problems long af, difficult to understand.
4. Lose interest, order dinner, watch BBT, eat.
Is it only me or others too that have this feeling?
+1 :p
you're not the only one
I guess BBT stands for Big Bang Theory. Anyway, this strategy proven to yield the best results during the last 2 div2 rounds...
*Rushes to the contest scoreboard upon reaches home
*Realizes how few people managed to solve Div1AB even though they appears to be totally solvable to me
*Feels bad about missing a chance to climb a lot of rank
*Realizes there are only 218 participants.
Whining asides, I wonder why so few people are participating the div1 round even though the time is completely fine, not to mention uni students are having a summer break.
Because A, B are extremely shitty, I guess
So this is a contest consists of only purely mathematical problems, it's fresh!
you're Right -_-
Writed solution for Div2B. TLE 7 test. Too much math. Even don't want debug it... Just want to know: is it bruteforce with ternary search?
I think brute-force + math is enough!
I fixed 2 points 1 and 2 and checked all the of the angle with third point. Say n = 5, so interior angle is 108 and increment in angle is 36 so possible values are 36,72,108. I dont know if it is correct
The angles are integral multiple of (180/n)*i where 1<=i<=n-2.
I think when you experiment with a few n-sided polygons then you realize that the total number of angles you can form with three points is n-2. Then you just iterate through all possible angles to get your optimal answer.
The idea in my mind is about: Calculate the angles of 0-1-(n-1), 0-1-(n-2) ... iterately by using sine & cosine theorem, then just binary search for the answer.
Say you are considering angle A-0-X with fixed A and unknown X, this angle equals to (0-1-2) — minus the angles at the two sides, which you can lookup and binary search from the preprocessed values, use binary search to find the two Xs which are just below & above the desired angle.
I think what you try to pair V[1], V[i] and to a binary search for the third one.
this my approach :
each angle = 180 * (n — 2) / n
you can divide each angle to (n-2) parts , the degree of each part = angle / # of parts
start withe one part and loop , each time add a part and optimize your ans.
O(N)
You can then set that angle to a and solve for n to get a O(1) solution.
It isn't much math, really. There's no need for ternary search. You have to consider the circumference the the polygon is inside, then the only observation you have to make is that after fixing the size of the arc, any point would suffice as midpoint. (consider a, b and c as points of the polygon. For simplicity, a=1. Also, a < b < c. Then, for every c you choose the angle formed by acb (in this order) is the same).
How to solve B:
First notice that any interior angle for a regular polygon is (n — 2) * 180.
Second note that if we take 2 1 as V1 and V2 we can see that we can iterate from 3 to n and try each one as v3 because N is small enough.
Now we see that the difference between 2 1 x and 2 1 (x + 1) is always the same which is (((n — 2) * 180) / (n — 2)).
So just look up all possible v3 and choose the best one.
I could not get why the difference is always same. Is there any reason or is it just an observation ?
Because of symmetry if the difference is not the same then this means that there are angles and/or sides that aren't equal which is not correct because the polygon is regular.
Never thought like this. Thanks
no problem.
:)
Let's assume there are three vertices, v1,v2 v4. There is another vertex v3 in between v2 and v4. Now, it is not difficult to prove that angle(v2v1v3)=angle(v3v1v4) when the edge connecting (v2,v3) and (v3,v4) are equal. Just join v2 with v4 and v1 with v3. Since the edges (v2,v3) and (v3,v4) are equal, then tiangle (v2,v3,v4) is isosceles and therefore (v1,v3) will pass through the mid-point of (v2,v4). Hence, the difference is equal.
Just realize the angle with Vj-1 Vi Vj+1 formula is equal to eachother, then i just do some stuff with some conditions.
It was bad idea to solve d before solving b. Hope my rating won't fall much.
Is div2C/div1A just a stupid case analysis, or is there an elegant solution?
Actually stupid case analysis is necessary as far as I understood.
You can divide cases with
b >= aanda > b, and again,r-l+1 >= periodandr-l+1 < periodwhereperiod = 2*(b+a)Consider these cases
a = 4, b = 2:abcd,dd,abef,ffrepeats in optimal answer, anda = 2, b = 2:ab,bb,ac,ccrepeats in optimal answer. In the first case, if the query interval is equal to or longer than the period, then answer is clearly2*a-b, and in the latter case, it'sa+1. If the length of query is shorter than the period, then we can just loop over the string which contains only one period.If there's nothing more than if/else in the solution, i think there are just too many cases.
You can exploit period for large cases, and brute-force(manually construct short — whose length is around 50 — string) for smaller cases.
I solved it by checking the length of the interval, and if the interval is small (say R-L <= 100) then I brute-force the solution with simulation, and if the interval is bigger, I just calculate the worst case and print that.
Case analysis and modulo?
For instance when b < a, I was thinking maybe the first 2*a+b characters can be repeated. Knowing this we can iterate from min(L%n, R%n) to max(L%n, R%n) where n=2*a+b. But I don't know if this is correct. :(
I did the same thing! Just be careful though, since there might be a case for example where n = 12, l = 9 and r = 14. You would be iterating from 2 -> 9, and this might give a WA.
No need for case analysis actually. I used bruteforce to compute first
12 * (a + b)characters (random constant) and then, ifr - lwas long enough — print number of different characters in string, otherwise modulolwith length of the pattern and count distinct characters in substring of right size.How to solve Div. 1 A C D E? :)
By your way to solve, I think you will ask about A B C D :)
Can someone give idea of Div2B ?
Check the angle between the diagonals starting from one verticle.
Yes I did the same.Plz correct me .
double angle=(double)((180)/(n)); double a1=angle; double m=INFINITY*1.0;
for(ll i=3;i<=n;i++) { if(m>(double)abs(a-angle)){ m=(double)abs(a-angle); ans=i; } angle+=a1; } cout<<"2 1 "<<ans;At calculating the angle, you do the cast wrongly. The program calculates ((180)/(n)) first, because of the parenthesis, and after casts this to double. But calculating ((180)/(n)) already floors the result to an integer, so casting it to double doesnt matter. Hence if e.g. n = 100, then angle should be 1.8 but it will be 1.0.
To fix you simply have to delete the parenthesis, and write "(double)(180)/(n)".
Thank You Bro! lost so much for small mistake!hope that i will not repeat it again.
Because of symmetry, you can fix the first vertex of the angle, and itterate the other 2 vertices. To avoid O(N^2) complexity, you can only itterate the second vertex, and find the optimal third vertex using binary-search. All you need is a function that calculates the angle of 3 given vertices in a regular n-gon.
Please can you once check the code above.Thank you
I think that we can even fix two adjacent vertices, because size of the angle can be calculated using number of arcs the angle takes.
I talk about arcs that n-gon divides the circle into, if we inscribe the ngon in a cirle
Yes. I just kept the first vertex at 1 and the second vertex at N. Then I just used some geometry to calculate vertex 3.
Actually you can do it in O(1) with some geometry.
Actually, you didn't even need to iterate the second vertex. All you needed was the third. If this doesn't hold true then RIP my rating :D
Yeah, i iterated only the third vertex too
How to solve Div2 D?
You can calculate dev[i] = deviation of permutation with i cyclic shifts. You can calculate it fast with prefix sums.
Can you elaborate little bit more?
Let's say p[i] >= i
In first p[i] — i cyclic shifts p[i] — i will be not negative.
And if p[i] < n next n — p[i] cyclic shifts p[i] — i will be negative.
And the remaining cyclic shifts p[i] — i not be negative again.
You can add arithmetic sequence in range with prefix sums.
You need to do the same think for p[i] < i. Hope it is correct.
Thanks for this nice approach!
Your approach is elegant :) Thanks!!
I had similar idea when I was trying to solve it. But my problem is: how can you do range update when you do not add same number? You need to add in range [x,y] lets say. But yu need to update p[x] with p[i] — k, p[x+1] with p[i] — k — 1... So you dont update with unique number. How do u handle that?
I think I got it. Range update with arithmetic series as said above.
I will be pleased if you explain how to use prefix sums to calculate dev[i]
Can anybody tell my I was failing on pretest 6? My code — https://pastebin.com/R09b5YRN
Problem is Div2B
Same Problem Bro!waiting to see the solution
After discussion with a friend, I guess I have done mistake in taking n as integer. Not sure about yours though.
How to solve Div 2D?
I tried but failed. Maybe try to use prefix sums? Here's my code:
Given a[], foreach a[i] — i > 0 push (a[i] — i + t) where t is the current time into a heap, for each iteration, update the value that has been newly pushed to the front, plus one for each element which is not in the heap, minus one for each elemnet which is not in the heap, pop elements from the heap if time = a[i] — i + t, as now a[i] — i should be >= 0.
Edit: It appears that O(nlogn) is not good enough, to achieve O(n), use prefix sum as others mention to store the updates instead of using a heap. The idea remains simular.
How to solve Div2 C?
I greedily generated string upto 2*(a+b) length such that Mr. B always choose the last character from the generated string so far. Ex — If the string so far is "abcde" then Mr. B will form string "eeee..." and merge it to the original string i.e "abcdeeee...". After generating the string I mapped the value of 'l' to smaller string by taking mod with (a+b). http://codeforces.me/contest/820/submission/28096854
I followed the same approach. But how did you assign the value to l. Why in particular mod with (a+b) and not the length of string. The optimal string have a length of 2(a+b) right?
You can observe that in any case after 2*(a+b) length, the string will get repeated i.e s=s+s. Also there is Symmetry in string from length (1 to a+b) and (a+b+1 to 2*a+2*b). You can also take mod with 2*(a+b) but then you have to increase the length of string to 4*(a+b).
How does the string have symmetry. As I understand, the length 2*(a+b) forms the period. Eg: a=2,b=2 Then string would be abbbaccc|abbbaccc|... (so on)
Sorry, correct me if I am wrong but can you explain in more detail, probably with an example.
What I actually meant was abbb and accc are not exactly same but they are similar. This similarity can be used and that's why I used (a+b) to take mod. But you can also use 2*(a+b) to take mod.
Thanks!
I didn't solve the problem at contest time, but I think that the first thing to do here is to split the problem in two cases:
if r — l + 1 >= 2 * (a + b)
in this case the person A is going to put a different letters for sure, since there is at least a segment of a letters in which person A is going to put his different letters. Since person B should play optimally then he's going to put the same letter and it should be one of the letters from the first A segment. but since there are 2 segments for A, then A cannot put the same letters as in the first segment since B already put one letter and A look for a suffix of at least 1.
So, in this case the solution is a + 1
else
simulation, starting from l to r starting with the letters from A or B depending on l % (a+b)
How to solve div2C ??
How tp solve Div2 C and D ?
I must be the only one solving Div2 Problem B by constructing a regular n convex polygon, and using binary search to find the answer :P Totally forget things about circles
I did BS then I realized it's 10^5
You can construct the polygon in O(n) and use binary search to check every vertice, so the overall complexity is O(n + nlogn)
Yeah, but all those angles then you do binary search are the same...
I did the same :D
is Div2C just about implementation?
Problem A is a pile of bullshit.
Why? I thought that it was interesting.
I don't know whether it's good or not, but I am sure that people won't like that task ))
UPD : It is also very bad lolololol
*the whole contest
Solution for A:
Edit: It fails like most passing solutions, unfortunately. Details below.
why 96? I think 48 is good.
Not always, take a = 5 and small b.
I used the exact same approach to solve this problem but with string period = 2*(a+b) for any value of a nd b (off course a and b within the given constraints).
Although repeating last character of string during Mister B's turn seems to be a wrong idea (proved by LLI_E_P_JI_O_K in comments below).
Let us assume a=5 and b=1, then
initial s=> "abcde"
after first turn s=> "abcdee"
after second turn s=> "abcdeeafghi"
after third turn s=> "abcdeeafghii"
after 4th turn s=> "abcdeeafghii" + "abcde"(same as intial string).
Your reasoning is perfectly fine, It seems I miscounted when I wrote the post (post updated).
But is that true? What about the test case from below?
Your solution would give:
But the real answer is:
Of course, it fails, like most passing solutions, unfortunately. There's an interesting psychological factor regarding this problem. A lot of yellows and reds would just find something intuitive enough to be correct and (maybe) try "false-proving" it. Then, there are also a handful of targets who could not find something provable / ruled the expected approach out by finding counterexamples. What's clear for sure is that it goes well beyond Div1 A, if it's solvable at all.
IMO, problem D (div 2) was pretty cool. Shifting the sequence is equivalent to subtracting/adding one from each element in the answer. There's actually an easy way to do this.
Code
Could you explain your solution a little more.
Assume we are shifting to the left instead of the right (that's how I did it in my code, you can easily switch between the two).
An important observation is that, when moving from one cyclic shift to the next, you have to consider for each i from 1 to n, if p[i] < i. If p[i] < i, then it will contribute -1 to the answer for next cyclic shift. Otherwise, it will contribute 1. This is because every i will decrease.
Now, we just need to find where p[i] becomes >= i using multiset/priority_queue, and handle the case where the frontmost element gets shifted to the back.
Instead of multiset/priority_queue, you can use an array and add +1 at index k (=when), resulting in O(n) instead of O(n log n). (Apart from that my solution is more messy than yours.)
Code: 28096290, variable "change"
Div2C Maybe I didnt understand statements, but why in samples computer moves first?
Third paragraph:
"Initially the players have a string s consisting of the first a English letters in alphabetical order (for example, if a = 5, then s equals to "abcde")."
Oh, of course. Thanks
That moment when you know your solution won't pass system test.
Too mach math problem and I don't like it. I only solve the div2 A D. I even forget The Law of Cosines to solve problem B
You don't need Law of Cosines to solve B.
What .. So how to solve B
You need to know, that the angle between diagonals starting from one verticle is equal. E. g. in a hexagon the angle is 120°, and 3 diagonals go out from one verticle, breaking the angle to 4*30°.
Нормально ли, что решение в Div1 B за NlogN получает TL? даже рукописный декарт не помог.
Кажется, оно предполагалось за O(N), судя по ограничениям)
Так и есть. И там довольно просто, без каких-либо алгоритмов.
NlogN = 1e6 * 20 = 2e7 должно заходить учитывая что TL 2 секунды
Вы знаете слишком много алгоритмов, это вам вредит
I know solutions to BCDE but not A. I kinda think swapping A and E makes a better round.
The constraints are small enough that you can just simulate reading every page. No need for some fancy solution :)
EDIT: I forgot that div1 exists, my mistake... :)
He is div1 not div2. He talks about your C problem.
I think he meant for div 1
I highly doubt he has bothered to even read Div2-A. He is talking about Div1-A, or Div2-C.
You just have to check cases. If the gap between l and r is too big, then the answer can be calculated just from a and b, if the gap is small enough, then you can generate the string, and simply check.
Seems that you also failed system test...
Yes I did :(
But the approach is correct, just there is a bug in my code. (I guess, I haven't found it)
Yes, the approach is correct.
I've just got accepted now, and I can't believe that the only bug that made me FST on A (on test 71) is that, I used 1~26 to represent a~z, but I checked if 0~26 appear in the suffix when adding letters by Mister B's opponent...
UPD: it turned out that the author's solution was wrong... so we can only say that the approach is correct if the author's solution was correct...
I agree. E is the easiest, I think.
I couldn't make it till the end of contest but here's my implementation of E, finds solution for all possible cases and I checked the answers so I think it works.
http://ideone.com/zzLnUp
You can understand the solution from the code.
What's the intended complexity for Div 1 C? Is something like
per test supposed to pass where ω(n) is the number of distinct prime factors of n and 
is the number of divisors of n?
By OEIS (see Highly Composite Numbers) we have σ(n) ≈ 105 in worst case so with T ≤ 1000 this should work. (I also have this complexity)
Indeed passed. :)
I don't see any submissions from this id!
Edit: ignore.
Thought div1 B can be solved in
, wrote a bug, fixed and submitted at 01:59:33, and found
Time limit exceeded on pretest 10:(People that write difficult algorithms in easy problems and don`t get OK have always surprised me.
thank you adedalic for your efforts to make this round. but let's face it : your problems sucks ... it's even worse than robinyu problems in round 419.
Codeforces 101: Always use a smurf account when you feel like picking up a fight .... Whoops I am not doing this correctly, gotta run.
you are just being really toxic...
This is why you never forget int/int doesn't magically turn into double.
Thanks, C++.
I cannot agree more
That moment that div1 E solution looks simpler than div1 A lol.
That moment when you get popups for announcements 35 minutes after contest end xD
I think this contest was more math contest, than programming contest.
Stupid question, Can Div.2 D be solved with ternary search???
I also thought about this one but it would not work anyway
determine in which id p[i] > i
this makes 3 segments at most for each i
then add the abs(p[i] — i) using above and some array 28099494
wrote O(NlogN) solution with
std::multisetfor Div2D 28098060time limit exceeded on pretest 7
started coding discretization + Fenwick tree
contest was over
waitted for system testing
submitted and accepted 28098915
People that write difficult algorithms in easy problems and don`t get OK have always surprised me.
Got AC with multiset. There are much less operations to it, though, than in your solution. I know it's easy to make O(n) of this but it still runs in 900 ms.
http://codeforces.me/contest/820/submission/28095453
it is unrated :(
The worst problem for C div.2 :|
Can someone explain the answer of testcase #24 of Div2 C? Input: 6 1 654321100 654321115 Answer Given: 11
It should be 7. String can be abcdefgabcdefgabcdefgabdefg...
this is wrong. initial string => "abcdef" Now let us assume Mister B appends character 'g' to it (although it is not optimal case)
after first turn string => "abcdefg"
now computer(other player) cannot append any of the last 6(according to your test) characters of the current string.
So after second turn string => "abcdefgahijkl"
and so on
Your answer is wrong :(
Sorry, adedalic, but problems were awful (IMHO).
Thank you for this contest adedalic, i enjoyed solving problem B even that i do not like geometry problem but this one was worth thinking on it and really good problem.
Also i enjoyed trying problem D even that i had overflow that i didn't pay attention to and still stuck in pretest 7.
thank you again and waiting for editorials.
Misread Div1A statements. Passed pretests.
This is strange. Of course, I should've read the statements more carefully, but there were at least 15 pretests, so I've expected them to be strong enough. But all of them can be thrown in the trash bin.
Hello!
Can anybody tell me, why for Div2 C problem for test 64 answer is 5? I suppose I have better answer — 4.
Test 64: 3 1 4 10
My string for answer 4: (divided by players moves) abc b ade d ab
Segment [4,10] contains only 4 different letters — a,b,e,d. Am I wrong?
Thanks a lot for possible explanation of this issue :)
If the first four letters are
abc b
Then the next set will be
acd
So the string will then be
abc b acd
which invalidates your idea.
It can't be "abc b acd" — because "acd" contains letter "c" which is in suffix "bcb" already
You're right, I read the computer as taking suffix of length
b, not of lengtha.Then it looks correct to me. I'm not sure what's wrong here.
Are you going to retest the problem? :) And what about rating questions then?
Maybe you should write a private message to adedalic and tell him about this case.
Yes, I've already done it, but he, Kan and Mike haven't answered yet
The computer will look at the string "bcb" of length a = 3, so he will append the string "ade". LLI_E_P_JI_O_K's answer is OK!
So, as you can understand I got WA test 64 on this problem :(
abc b acd d abc ?
[4;10] there are 4 different letters
Yes, and answer should be 4 instead of "5" — jury answer.
Please, correct test 64, run total retest on this task.
same here! my program got answer 4 with abc b ade d ab ... too and i think my answer is correct!
As I mentioned above, the computer will append "ade" on its first turn.
Yeah, i am wrong
abc b ade d abc
but [4;10] also there are 4 different letters
This is exactly LLI_E_P_JI_O_K's answer from this comment :)
well wrong testcase answer.. but still nice problems..:)
This round had not 1 or 2, but 4(FOUR) testers. And none of them implemented bruteforce solution for A? I've always thought this is exactly testers' purpose.
I have implemented brute force during the contest, where the number of segments A or B is less than 3, for other cases got an answer by formula, as you can see, there are some bugs in tests :))
As I thought, it's setters task on codeforces to code naive solutions, and I'm always doing it when setting an own round. As I've understood, testers should just solve the tasks and estimate the difficulty of the tasks...
By the way I hadnt solved this task myself...
dotorya : Now I don't know how to solve A
So....
:<
http://codeforces.me/blog/entry/52941#comment-369823
Ops, my rating stays same :D
I got 1st place in Div2 and CF-predictor tells me that i'll have +310..
but I think this round should be unrated :P
( I just implemented C without any formal proof, and I knew some cases like this will happen...)
We're looking into the issue.
Beware, if you look too long into the issue, the issue will start to look into the you.
I have to use this sometimes.
There are fewer than 10 people in div1 having WA on test 64 (and they're mostly purple). If pretests hold, and there isn't any hack using such test case (why would there be, when the answer is wrong?), this might be the least solved div1A ever!
Yes, of course, I suppose Div2C too :)
Which kind of makes me understand why the testers did not spot the flaw. Because almost everyone fell for the same trap in the real contest.
Yes, but they should do some kind of slow but correct solutions to test small testcases and finding bugs in really fast and intended solutions:) But they didn't do that for this problem, I guess :)
I can imagine that it seemed like an obvious solution (it didn't to me though), and they might even have some proof with a tiny flaw somewhere.
It happens. I wish the author(s) to shake it off. They are way more invested in this round than we are. And it's not like someone tried to pull a notorious coincidence on us.
Well, we do have bruteforce solution, it produces correct answer on your testcase. We also run a stress test with it several times, it didn't find any tests against the author's solution :( And this test was a hack.
Right now we're looking if there is a solution to the problem or not.
Ok, thanks. If it is needed I can explain my solution of how to solve div2C/div1A
Ok, please explain if you think it's correct.
Ok, I'm going to explain. It will take some time, about 10-15 minutes.
I write from mobile, sorry, if something looks badly. Let's call string of second player move — "B-string", string of first player move — "A-string". First of all I have found how many times A-strings and B-strings appeared in the given segment L...R, with some simple math tricks and partial sum principle like F(R)-F(L-1). Then if B >= A, then every time on A player move total suffix will be inside B string, it's obvious that here we should make B-string with one repeated letter and if the test is quite big (I.e. we have >= 3 full A-strings) the answer will be always (1+a). When B < A, and count of full A-strings is >= 3, every such full string will give |A — B| new letters comparing with last A-string, but such A-strings can be repeated in chess-order, like. A1, B, A2, B, A1, B,... and B should consist of one letter from A1, then one letter from A2, then from A1,... Like in 64th test, but here we have less full A-strings, than 3. In such big cases (>=3 A-strings) answer will be minimal and equal to 2*A-B. Small cases I solved with bruteforce, small — when count of full A-strings is 2 or less, here the total length of segment is short and we can use bruteforce for B-string letters.
Tests with small count of covered A and B segments are quite tricky to implement through formulas, that's why I have preferred to write brueforce instead :)
=========================================
For big lengths seems correct to me (for b > = a the answer is at least a + 1, and the construction provides such a strategy. for b < a the answer is at least 2a - b, and again the construction seems correct).
However, for small lengths, it's not obvious that we should use only one letter, and that the answer depends only on the moves we make close the the segment [l, r].
Yes, but all letters from A are still different, that's why if we take such small segment, but L and R is too big, it doesn't matter what previous A strings were if we take B string consisted of one repeate letter, the same count of different used letters for pair ("previous A ans", "previous B ans") still be provided before the next A string. So, if we use strategy with all B answers consisting of repeating letters (possibly different) — nothing changes for us, we can look only on our segment. Isn't it?
And is that any idea of using B answers of different mixed letters? It seems not to be better or useful just because they will prohibit more letters from the next A string. Seems to be a proof somehow.
Quality of the problems is not good enough, one problem still without a correct solution, it has been about four hours since the contest is over, do we need more reasons to make the round unrated???
Problems were fine,I liked Div2 D and Div2 B,both of them could be solved in couple lines of code,very nice problems.If problem div1 A hasn't got solution it will definitely be unrated.Main problem is that 2 consecutive unrated rounds will make joke out of codeforces.
And what about the round's 421 destiny, this decision will be rather difficult, in every case, I suppose, some contestants will be slightly frustrated.
I would be furios.I made a bet with my friend from school that I will be div 1 on 1.st September and I really practised a lot.Last 2 contests I got over +170 in sum and if both of them end unrated,I will just give him 20 euros and stick with atcoder rounds.
Ummmmm, but you didn't even submit Problem C...
I tried to solve it and had something but couldn't prove it.The point is that I would have over 1600 rating if last 2 round were rated.I won't leave codeforces ,will do problemset because it helps to improve but I won't do rounds anymore.
That is unfortunate, but you have the right mindset in your last sentence. The goal is to improve. Your rating will follow. I would've just made div1 if last contest were rated, but it doesn't bother me that much. What I do care about is whether or not I'm improving and it looks like we both are. Keep practicing and you'll make it. You have plenty of time. Best of luck to you!
Thanks man,best luck to you too!
After all that's happened, I'm curious:
How many hack cases are added to the main tests? How is it decided which ones will be added?
I'm guessing there must be an upper limit and an algorithm which picks random hack cases, but I couldn't find any blog which mentioned so
Usually the number of hacks is not so big, so we add all of them. Even if there are 100 hacks on some problem, probably most of them are equal, and only distinct tests are added.
Here is my submission, can you check it? http://codeforces.me/contest/819/submission/28109245
I think it's so simple it's like how I solved the problem: abc d abc d abc d abc ans = 4
And you are not right,because the second string ABC can't contain letters B and C, they are in suffix Bcd
Wow, the test is made by me...
Amazing intuition! My congratulations! :)) Thanks a lot for this tricky testcase :)
I think problems were fine ! Just estimation of level is very bad + bad pretests for A div 1 :)
Anyway I am stupid guy and I should stop with competitions.
Hi guys! Something bad happens to cf-predictor, it's falling over and over again. It's not clear what is the reason. You could download prediction here. Mirror will be available soon.
UPD
I can see rating prediction on standings page, hope you too:) Website currently deployed on AWS
i think it's because cf-predictor didn't Predict that this round will be this awful
Stop flooding just because you didn't like the contest, please.
How about Div2.
Why do people put geometry in their problems, i mean it's clear math, informatics it's about implimentation and other informatics' topics. Already 2 contests in a row there is geometry, that's shit.
There is no royal road to Geometry.
Yes
you need to watch kevinsogo's video lectures
thanks i will try.
Can you please give a link to any lecture?
you can find it on youtube.
Search for geometry and number theory by kevin charles atienza.
Okay thank you
What lectures are you talking about? Could you provide us with a link?
Geometry and Number Theory By Kevin Charles Atienza | Indian Programming Camp 2016
Also, here.
I think the contest problem that it had depended overly on Maths and Geometry
adedalic Does the dataset of C has some kind of problem? My AC code performs the following: input: 4 2 11 21 output: 6
but it can be 5. The string could be like this: abcdddabefeeabcdddabef........
from 11 to 21 we can have only 5 distinct letters.
Please see to this if I have misunderstood something......
(I skipped my first code because I found this test case and submitted a new one... now I see my skipped code gets AC. This is the greatest irony)
Something is wrong with Div2C = Div1A systests.
In this comment there is information about a wrong test in the testset.
could you please see the testcase I have given? I had a hectic day and I may be wrong :v please let me know :v I need to know i haven't lost my mind totally
Seems correct:
Input:
4 2 11 21And
s[10:21]has only 5 distinct letters: a, b, c, d and e.See here, my solution has better ans than judge. Getting WA on this case. :(
Is it a wrong answer in the pretests? If so, the round must be unrated. Do you have a string which can be generated by the players and the answer for which is 12?
It should be rated if pretests were ok.
It should be unrated if there is no correct solution to that problem.
But I think someone have got the correct solution of this problem.
Not necessarily. We know that some solutions can generate a better answer. It's not known if these solutions are correct. There might be other test cases where the answer is even better than the one produced by these solutions.
Why this should be rated if pretests were ok ?? Someone may pass pretest and get stuck because judge data is wrong. @gabrielsimoes
Well, because even though system tests are wrong, system tests are not available until after the contest and shouldn't be taken into account.
You can use wrong solution to pass some of the codeforces probem's pretests, but it can be hacked or FST.
If pretests were ok and there is a correct solution to the problem, the can remake the data and rejudge the problem.
Ok, maybe, but I don't think it will be normal, if problem Div 1 A will be solved by 10 people, as all passed solutions are incorrect.
Here is the string for test 12.
Seems like this string is incorrect: on the computer's first turn it will look at last a = 12 symbols (
"bcdefghijkll") and append the string"amnopqrstuvw".Oh sorry ! My bad ! Missed it.
You solution produces an invalid answer. The piece of string
[l - len, l + len * 3)looks like this:"ghijkmmabcdefghijkllabcdefghijkmm". That's not a valid string.bcan't be there as it's exactly the twelve-th character to the left from its second occurrence.Rating Predictor ....?
in Div2 B test 6 the input was 100000 1 my answer was 1 2 181 the jury's answer was 2 1 558 can anyone explain to me why the jury's answer was better ?
Why don`t you make a polygon of 100 thousand sides for this?
Then just measure your angle with a protractor and jury`s answer also. I guess then everything will be fine.
Your answer is 1 2 181. The angle vertex is 2, and the angle it yields is huge (about 180 degrees). The jury's answer is 2 1 558. The angle vertex is 1, and the angle is yields is very close to a = 1 (about 0.999 degrees). Thus the jury's answer is better.
Try drawing the angles on this simplified 100 side polygon if you don't get my explanation.
thanks i got it now
The angle made by 1 2 181 is 180- 0.324 degrees =179.676 while the angle made by 2 1 558 is 1.0044 degrees.Now tell which has a lesser absolute difference ?
yeah i got it now i think it is a problem in the logic i used
Don't compare your answer with jury's answer, because there is more than one.
It can WA for some decimal digits.
How can we confirm then?
long double instead of int.
Here is my code. Can you tell me my mistake? https://pastebin.com/2jrLBGa8
Very long and hard to understand solution.
Hope this will help you.
It's not rated???
It is
.
for div2E or div1C am I correct to say that the the following block will give answer:
Not saying it will be fast enough just want to know if this is equivalent to the problem
That doesn't pass the sample. But the second line is correct (this accounts for the second beacon). And that part is quite simple to calculate fast.
Edit: Nevermind, you're right. I swapped N with M. This is not my day :-D
Really? I've coded it and it worked (this brute force approach which of course I didn't submit)
see edit :)
Ohh, now I see. Sorry
UPD:wrong
Yeah, this is correct (more or less, not really sure about +1 -1's)
For Div2B,I see solutions that fix the first 2 vertices and iterate for the third vertex assuming the angles thus formed are a multiple of
(180/n). Can anyone provide with the proof for this? I tried working out the values but I found that they were not always equal(I may be wrong :P).Let us first draw a circumcirlce of the given polygon.Consider any angle (v1,v2,v3).This is half of the angle formed by the chord (v1,v3) at the centre. (v1,centre,v3).So,given any chord you can easily determine the angle it will form with a point on the circumference.Let's start with the chord (n,1).The angle it makes at the centre is (360/n).So,the angle at the circumference will be 180/n.Move on to (n-1,1) : angle at the centre will be 360/n.So,each time 180/n gets added to angle being formed.
Nice Solution :)
:p
Check out my comment here: http://codeforces.me/blog/entry/52927?#comment-369764
You can draw geometric figures and verify the angles. They come up to be the same.
Downvote a contest first time. Now i feel sorry to my friend, who i invited to participate the contest. Obviously，the awful problems wasted his time. Not only div1a.
Do it unrated. :c
It seems the organization is still causing troubles in this time line.
There can't be two consecutive cf rounds unrated. send_nodes, lets send another D-mail.
That moment when you and send_nodes are in the same room. xD
I was there too. I tried to hack him, for the previous round where I didn't get +116 rating. :D
When you try so hard to improve your rating, but the round ends up unrated.
Somebody clarify rated or not?Its been 3 hours!!!
We are working on the issue with problem A in div. 1. We will announce the decision later. - ( The update in the blog )
They should give that hacker kudos for hacking the entire contest xD instant 1st place or something lul
Who do u mean?
A special position to him for hacking all hacks.
Can anyone help me find out what I'm doing wrong for Div2D? I don't know how to get past pretest 9. Code
just push it!
Oh !!! It works :)
Interestingly, Div2 B is educational and enjoyable. There are 2 solutions that come to my mind (assuming one of the outer points of the angle is fixed as vertex 1):
Iterate the middle vertex of the angle. Then, either binary search or use a two pointer approach to find the third vertex.
Remember your geometry theorems: The locus of middle points for the angle given the outer points are fixed is a circle (the circumcircle of the polygon). This way, there are only O(N) candidates for the last point of the angle and there is even a way to solve the whole problem in constant time by some formulas.
In this particular problem, I have a doubt- Why solutions are always such that two vertices are adjacent ? I mean in a 20-gon polygon why cant there be an answer like 4 7 15 for a specific value of angle a.
There are multiple solutions, but you can print any of them. Finding x so that the answer is "2 1 x" is the most practical approach.
it is not necessary as for a n=6 and angle=150 one of the possible answers is 1 3 5
Okay..But how is it possible that everytime this two adjacent vertex approach is validated ? Any proofs for that ?
This is how this work. Suppose your solution is angle 10,20,30. But, angle 11,20,31 is also a solution (because angle 10,20,30 is same as 11,20,31 -> you just "shifted" angle. Why? You can draw a circle around polygon. Then, every peripheral angle over same chord is equal). By same logic, angle 12,20,32 is same. If you go on like that, angle 19,20,39 is same as 10,20,30. So, whatever angle is your solution, you can shift it to angle that has two adjacent vertices. Also, its logical because everything is simmetric.
I'm not strong in math, so I coded a stupid solution with problems about precision, but it passed! http://ideone.com/LxRrwC
Two unrated rounds in a row.
I feel sad for those who got hundreds of down-votes because they asked if the round was rated or not :v
This contest was definitely unlike any other. The questions involved much more math than normal. In Division 2, the best way to solve Problem B was by drawing out a diagram. It could be realized that since a circle could be inscribed around the regular polygon, and due to the inscribed angle theorem we could just pick two points right next to each other for one side and test out the remaining possible points for the third point. That allows for an O(n) complexity and much easier implementation. Also, Problem D in Division 2 (also Problem B in Division 1) was surprisingly easy! If you haven't already, you should definitely check it out! Just watch out for time complexities!
Can you atleast give a rough idea of when the decision will be announced? I've been checking this page for around 3-4 hours now. :(
I believe that once a solution to div1 A/div2 C is verified, they will just rejudge submissions and provide new standings based on which, the rating will change.
You can speed it up by creating a formal proof of the solution to the above problem.
I think that would be best for now to rejudge in case of such an ambiguity....two consecutive unrated contest is not fine for legendary Codeforces
Upvotes for contest came down from around 250 to 50. If they don't announce their decision for some more time, it can come down to 0.
And maybe under!
This thread has more than 300 upvotes couple hours ago.. Now it's only 50-ish and falling..
Edit: it's negative now. Feel sorry for him. Hope this problem is solved soon before it gets worse
Div2C/Div1A
I am getting wa for below test case. My answer is 4 and jury's answer is 5. However I am able to get a string which gives 4 as answer. What am I missing?
Test Case: 3 1 4 10
sample string with answer 4: a b c b a d e e a b
you are missing this
You're missing the last three hours of shitstorm in this blog.
The judge solution is wrong, yours is (probably) correct.
My bad! Its a really long thread and I got little selfish!
Now this blog has negative votes...
And author's contribution will hit rock bottom soon. :-(
I feel sorry for him.
Is it rated?
So, now is the appropriate time to ask this question.
I'm sorry to say that the round is unrated now. Here are some details.
If the rating of this competition was a stock, then we would all be broke.
I would like to change my vote for this topic, however cf does not let me do this. Is that option blocked on purpose? I think that I should be able to remove my vote or change it any time.
I can't understand why all those downvotes there is always a possiblity for such a situation and the testers & the problem setter described that they did their best. I think we can understand the situation more wisely ..
I gave upvote and will always do so,people need to consider how much time authors putted on testing and making problems,mistakes can happen.On the other hand you must consider amount of unlucky people who performed for over +150 rating in previos two rounds and ended up at same rating.it's so frustrating,I can't go over 1480 and with these 2 contests I would be over 1600.Anyway after all this happened suddenly I don't care a bit about rating and won't participate in CF rounds for some time.
I'm really disappointed about the blog post rating is negative.
Several hour ago, the post rating was 300-400, but now it is -50.
I think some people was disappointed about unrated. Me too. But the contest writer worked very much for codeforces and people, and you may enjoyed the other problems.
So I think this post should be positive, at least 0 because you may enjoyed at least a little, and the writer worked very much. Don't blame the writer too much.
UPD: I see the blog again after 9 hours writing the comment, and the post rating was -18. Thank you for the incident didn't got worse, but the post rating is still negative. I say again: don't blame the writer too much.
UPD2: the post rating became to positive! I'm very happy. (The writer and most people may be happy too.) Thank you!!!
Man, it is just a rating, it is not like we are beating him up or something.
I belive this is a sign to stop the poor proof contest and the over confidence that the algorithm is correct because the problem looks simple.
I saw many comment of feedback of Codeforces Round #421, and I posted my opinion:
Yes I agree, writer prepared very much. And I think the other problems except Div2C/Div1A is good. Though I have came across similar issue in Codeforces Round #382, that is, there was an issue in Div1A/Div2C, and this is the similar pattern.
I agree that I think writer/tester should make a correct solution with correct proof, but I think Codeforces has too much rounds like 2-3 days per a contest. It is very very short period compared to Topcoder or Atcoder. I want to ask whether Mike or Kalinin can do really good testing and checking proof and checking the editorials. Though, we will be very happy if there's a contest per 2-3 days and also no issue with very high probability.
UPD: The contest announcement blog is now positive like +20. I also read E869120's comment below. I think many people don't have experiences of problem writing in a large-scale or international competition, and some of them may thinking that it is not very difficult. (Arpa's blog describes how Codeforces problemsetting is doing like.)
Thanks very much for upvoting the announcement blog.
Finally, the topic rating became to positive (Now it's +18!). I'm very happy.
I think any real announcement of CF should be positive, whether the contest was rated (or unrated), whether the contest delayed, whether the contest has some issues, whether your score or rank in the contest, and whether there is some notorious coincidences in the contest.
Please consider about the work of writer. It may took so long time, and writers/testers have many difficult tasks. I already hosted 4 contest which is unofficial, and 1 contest which is official, all of these in AtCoder, but in one contest example, it spent all of 2 weeks to prepare contest.
I hope there is no real announcement post of CF contest which has negative rating.
The test (Div1 A Test 64) is made by me.
Because of my test the contest is unrated, today is so cool for me XDDDDDDDD
Amazing job! :)
Achievement Unlocked:
Hack the contest so that make it unratedLast two contests,
Rated contest
Solve and hack
Sorry, it is unrated
First time I get rank < 50, contest is unrated :(
Lol, this guy didn't even bother to hide his second account. What an outstanding perfomance from the one who solved the most problems on codeforces ever!
Submissions are obviously the same. Ali.Pi's B, Ali.P's B.
Please no notorious coincidence memeВ конце написано Рабор должен быть
edit:I know now why it will not work :-( adding -1 will not solve the problem :-D
I have a solution for div2/C div1/A , I want to know if it will work , because it seems simple to me and I checked it many times and it worked
I considered segment [1..inf] (from 1 to infinity) that describes the "game turns" as repeated segments of length 2(a+b) (see the figure for the different situations)
I replaced the choice of Mister B by -1
The solution will be as following :adedalic can you take a look
my solution code
photo hosting service
problem writes are just pieces of shit.