By PikMike, history, 9 months ago, translation, ,

Hello Codeforces!

Series of Educational Rounds continue being held as Harbour.Space University initiative! You can read the details about the cooperation between Harbour.Space University and Codeforces in the blog post.

This round will be rated for the participants with rating lower than 2100. It will be held on extented ACM ICPC rules. The penalty for each incorrect submission until the submission with a full solution is 10 minutes. After the end of the contest you will have 12 hours to hack any solution you want. You will have access to copy any solution and test it locally.

You will be given 7 problems and 2 hours to solve them.

The problems were invented and prepared by Roman Roms Glazov, Adilbek adedalic Dalabaev, Vladimir Vovuh Petrov, Ivan BledDest Androsov and me.

Good luck to all participants!

UPD: You can discuss the problems after the contest on a local Discord server.

Our friends at Harbour.Space also have a message for you:

Hello Codeforces!

A quick reminder to check out the Hello Muscat Programming Bootcamp!

Every day, both the Hello Programming Bootcamp and the Moscow ICPC Workshop will be competing simultaneously, 4,000 kilometers from each other — our camp will take place in Oman, from March 9th to March 15th, 2019 — registration is still open for this unique opportunity, so be sure to give it a look, we would love to see you there with us!

REGISTER FOR THE BOOTCAMP

And, don’t forget that Harbour.Space University’s opportunity for a fully funded Master’s in Robotics programme scholarship is still open! Head over to their website to see the details, and apply before the registration date closes.

APPLY HERE

Congratulations to the winners:

Rank Competitor Problems Solved Penalty
1 V--gLaSsH0ldEr593--V 7 198
3 Rzepa 7 396
4 eddy1021 6 170
5 spacewanker 6 175

Congratulations to the best hackers:

Rank Competitor Hack Count
1 _bacali 39:-44
2 Kiri8128 8
3 tymo 5:-1
4 CoodeForces 4
5 Sevlll 5:-4
121 successful hacks and 527 unsuccessful hacks were made in total!

And finally people who were the first to solve each problem:

Problem Competitor Penalty
A V--gLaSsH0ldEr593--V 0:01
B neal 0:02
C neal 0:05
D V--gLaSsH0ldEr593--V 0:10
F V--gLaSsH0ldEr593--V 0:44
G Kalam 0:17

UPD2: The editorial is out

• +75

 » 9 months ago, # |   +20 Clashes with Topcoder SRM. Please prepone it by 30 minutes.
•  » » 9 months ago, # ^ |   +30 Also clashes with CodeChef Lunchtime.
•  » » » 9 months ago, # ^ |   +20 It's surprising that two contests promoting the same bootcamp are clashing.
 » 9 months ago, # |   -15 clashing with codechef lunchtime please delay it.
 » 9 months ago, # |   -84 Clashes with indian entertainments
 » 9 months ago, # |   +89 Clashes with Educational 59
 » 9 months ago, # |   +12 中国玩家修仙场：）
 » 9 months ago, # |   -13 Clashes with CodeChef lunchtime. Please reschedule.
 » 9 months ago, # |   +17 Happy Republic Day Indian Coders!
•  » » 9 months ago, # ^ |   -6 thanks!!
•  » » 9 months ago, # ^ |   -28 like who cares to be honest?
•  » » » 9 months ago, # ^ |   -11 All Indians in this Codeforces Community! ;)
•  » » » » 9 months ago, # ^ |   -26 and who cares about them?
•  » » » » » 9 months ago, # ^ |   -12 Codeforces ! and you don't have to!
•  » » » » » » 9 months ago, # ^ |   -28 i strongly believe they don't.
 » 9 months ago, # |   +3 Please make contests at more diverse time frames -- times like 18:35 MSK are really unfriendly for students living around UTC +8.
 » 9 months ago, # |   0 PictureHope every one become candidate master one day.
•  » » 9 months ago, # ^ |   0 nice meme bro
 » 9 months ago, # |   -10 Hope that will be a lot of hack today xD
•  » » 9 months ago, # ^ |   -7 I try to hack the problem B&&D , but I failed
•  » » » 9 months ago, # ^ |   -7 they have made good pretests today
•  » » » » 9 months ago, # ^ |   -7 They have made bad problems too
•  » » » » » 9 months ago, # ^ |   -7 yeah that's for sure xD
 » 9 months ago, # |   0 Nice picture though~ :D Is that the campus?
 » 9 months ago, # |   +1 Give me a purple name plz...
 » 9 months ago, # |   0 It shows the possibility of being a good hack :D, hope every one of us succeed!
 » 9 months ago, # |   0 Want to Be a Candidate Master... So everyone Good Luck && Higher rating... RP++
 » 9 months ago, # |   +40 Clashes with clans
 » 9 months ago, # |   -36 oh my the name hints it will have pointless math again. god, can't you keep these two things sepparate? is it so hard to understand that math has nothing to do with programming? it doesn't help you get better, it's just useless.only people who want to feel better for themselves thinking they've actually accomplished something like math in programming.so with this said let's hope for a math-free problem set.if it doesn't have any math i will surely get out of expert. can't wait.
•  » » 9 months ago, # ^ |   -15 last minute smart move: i'm not going to take part in this contest because i know from experience it's going to be a lot of math. not worth taking the risk.
 » 9 months ago, # |   0 CodeForces OR CodeChef???Both.
 » 9 months ago, # | ← Rev. 2 →   -30 Fuck problem statements. I'm done.
•  » » 9 months ago, # ^ |   0 I don't know why, but at last contests problems are easy or hard, there is no medium ones.
•  » » » 9 months ago, # ^ |   -33 they are math problems which are useless. it makes them pointlessly harder, that's the reason why. don't bother...
 » 9 months ago, # |   0 For the question D, if i=1 and x=4 then [i/x] gives 0. This index does not belong to the matrix B What is to be done then ?
•  » » 9 months ago, # ^ |   0 There is ceil function, so ceil(i/x) gives 1.
•  » » 9 months ago, # ^ |   +2
 » 9 months ago, # |   +9 nice problems ^^ , how to solve E ?
•  » » 9 months ago, # ^ | ← Rev. 3 →   +1 https://uva.onlinejudge.org/index.php?option=onlinejudge&page=show_problem&problem=1500Almost the same logic XD
•  » » » 9 months ago, # ^ |   0 What's the DP state though?
•  » » » 9 months ago, # ^ |   +18 What is the solution for this one? The constraints here suggest that n3 is possible.
•  » » » » 9 months ago, # ^ |   +13 For the UVa problem in the link: Let's calculate dp[l][r][len] — the most points we can gain if we do operations on s[l] to s[r] with the same len characters as s[r] attached at the back of the string. Let p be the leftmost index satisfing s[p] = s[p + 1] = ... = s[r] and q be some indicies satisfing s[q] = s[r] and s[q] ≠ s[q + 1].Then we can list the dp equations: dp[l][r][len] = dp[l][p - 1][0] + (r - p + 1 + len)2 — if we clean all blocks from p to r ; dp[l][r][len] = dp[q + 1][p - 1][0] + dp[l][q][r - p + len + 1] — if we decide to clean blocks from q + 1 to p - 1 first and the others at last. For today's E problem just substitute (r - p + 1 + len)2 with the most points we can gain from the string with same characters of length r - p + 1 + len. This can be done by another simple dp.I know the constraints suggest a O(n3) solution, but maybe many states cannot be approached? I desire the proof too ><
 » 9 months ago, # |   0 contest was so easy :D
 » 9 months ago, # |   0 How to solve D ?
•  » » 9 months ago, # ^ |   +7 Mine's using sliding window. Keep extend our window when the string in the next row is equal to the string in the current row. Keep track all size of the windows and find the GCD of all windows' size
•  » » » 9 months ago, # ^ |   +3 Nice idea, but I think you should do the same thing on columns too.
•  » » 9 months ago, # ^ | ← Rev. 2 →   +9 Basically we need to check over all x such that x is factor of n,for a given x we need to check all the [n/x]*[n/x] sub-matrix whether they contains equal element or not, if a submatrix contain equals elements either all of them equal to 0 or 1 i.e sum is either 0 or x*x,we can easily check sum of a submarix with the help of 2d-prefix sum.
•  » » » 9 months ago, # ^ |   +6 Thats O(n^2*sqrt(n)) in worst case?
•  » » » » 9 months ago, # ^ |   +5 Nope, a number can have max of n^1/3 factors.and for each factor we are doing only n^2/x^2 work so actual complexity would be near to n^2 with some constant.
•  » » » » » 9 months ago, # ^ | ← Rev. 2 →   +14 n^2(1/1^2+1/2^2+1/3^2+...+1/n^2) < n^2*(pi^2/6)https://en.wikipedia.org/wiki/Basel_problem
•  » » » » » » 9 months ago, # ^ |   0 It is even less because you only check the devisors not every number.
•  » » » » 9 months ago, # ^ |   0 if you just check from all the numbers from 1 to sqrt(n) for a divisor of n, then this is the complexity in the worst case.
•  » » 9 months ago, # ^ |   0 Don't know if this is the indended solution, but I did this:Convert hex to binary and store it into a matrix. Then you try each divisor of N as x in O(N*N). At most, N has 60 divisors (N = 5040). I've executed this solution with the worst case in codeforces custom test and it seems to pass, but who knows...
•  » » » 9 months ago, # ^ |   +1 What is worst case in your opinion?
•  » » » » 9 months ago, # ^ |   +14 You are trying to hack me haha. Find it out yourself
•  » » » » » 9 months ago, # ^ |   0 I think that 5040 and whole plane in F will hack you, however im not doing it. If im not mistaken you gonna do ~25kk*60 operations.
•  » » » » » » 9 months ago, # ^ |   0 I've tested that case during contest, nothing happened, no TLE. Was thinking in solving it with binary search but that failed so i resorted to this.
•  » » » » » » » 9 months ago, # ^ |   +3 How is it not giving TLE when doing > 1.5*10^9 operations + big input?
•  » » » » » » » » 9 months ago, # ^ | ← Rev. 2 →   -8 You forgot that checking big x is not O(N*N). It is O(N*N/(X*X)). So the solutuon is way faster. Upd: I'm sorry I haven't checked your submission. Yes, your solution runs in O(N*N*Number of divisors)
•  » » » » » » » » » 9 months ago, # ^ |   +3 My code checks every x in O(n*n)
•  » » » » » » » » » 9 months ago, # ^ |   0 So how is it doing 1.5*10^9 operations in time limit?
•  » » » » » » » » » 9 months ago, # ^ |   0 I've seen an O(n) simple loop when N = 10^9 and it executed in 0.7s. It seems I do a ton of operations but most of them are just comparisons
•  » » » » » » 9 months ago, # ^ | ← Rev. 2 →   0 For that to work, the time limit of the problem should have been 1 sec. The solution having complexity O(N*N*60) gives 1578ms and the one with prefix sums (optimised) gives 400ms. The problem setter could have given it some thought and made it a better problem.
•  » » » » » » » 9 months ago, # ^ |   0 Mine runs in 0.7shttps://codeforces.com/contest/1107/submission/49014233
 » 9 months ago, # |   0 How to solve D?
•  » » 9 months ago, # ^ |   0 Find all divisors of n. Then for each divisor, divide the matrix into the dXd submatrices and check all the submatrix has similar elements (1 or 0). Find the maximum divisor that satisfies this.
•  » » » 9 months ago, # ^ |   0 I have used binary search to find maximum divisor which satisfies x-compression,but it fails on test 4,can u explain why? if linear search is used to find maximum divisor then it gives AC.
•  » » » » 9 months ago, # ^ |   +1 Binary search wouldn't work as it is not a monotonous sequence.Eg:  12 E00 E00 E00 E00 E00 E00 E00 E00 E00 E00 E00 E00  The answer to this is 3, and both 2 and 4 don't satisfy as an answer.
•  » » » » » 9 months ago, # ^ |   0 OK Thanks
•  » » » » » 9 months ago, # ^ |   0 hahau
•  » » » » 9 months ago, # ^ |   +1 I also did the same, but here our assumption for binary search to work is that if the current_divisor is not answer then the answer is less than the current_divisor but this is false answer can be greater than current_divisor.
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9 months ago, # ^ |
Rev. 3   0

## if this is the case then the case 1 given should give ans "2"..2 x 2 matrix is having all equal elemts and 2 divides 8 too...so why the ans is 1?CAN U EXPLAIN THIS??

1. 11100111 E7
2. 11100111 E7
3. 11100111 E7
4. 00000000 00
5. 00000000 00
6. 11100111 E7
7. 11100111 E7
8. 11100111 E7
•  » » » » 9 months ago, # ^ | ← Rev. 3 →   0 Consider first two submatrices starting from the first row:1111and1010Here the second submatrix doesn't have similar elements (combination of 0 and 1). So 2- compression isn't possible
•  » » » » » 9 months ago, # ^ | ← Rev. 2 →   0 means we have to check all possible sequencely (n/x)*(n/x) matrix ??..i thought the first one accrdng to ur ans...i am confuse in submatrices...which sub matrices we have to take??
•  » » » » » » 9 months ago, # ^ |   0 Yes, you have to check all sequential submatrices.
•  » » » » » » » 9 months ago, # ^ | ← Rev. 2 →   0 thanks.
 » 9 months ago, # |   +4 what was the state for the DP in problem E?
•  » » 9 months ago, # ^ |   0 I think, (sum_of_current_length, left_index, right_index), but I'm little scary about time limit of my solution
 » 9 months ago, # |   +3 Is D solved in O(no.of divisors*n*(n/4))
•  » » 9 months ago, # ^ | ← Rev. 2 →   +5 It's
•  » » » 9 months ago, # ^ |   +11 My algorithm solves it in O(n^2). Idea: count the prefix sums on the matrix in O(n^2). Then, for all divisors (div) split the matrix in small squares of length div and check if it each of them has sum div**2 or 0. Each iteration works in (n / div) ** 2, but we should take into consideration that however many divisors of n there are, the resulting number of iterations must converge at appr. 1.5 * n^2 (n^2/1 + n^2/4 + n^2/9 + ...), therefore the time complexity is O(n^2). Correct me if I am wrong.
•  » » 9 months ago, # ^ |   0 mine is O(n*n*sqrt(n))
•  » » 9 months ago, # ^ |   +6 I have a much neater solution. Find all lengths of consecutive segments with same character. (like 0001111110 then you count 3,6,1) Do this for both vertical and horizontal. Answer is the gcd of all the lengths. O(N^2), because you can just determine whether a segment of this length exists, and so gcd would be called N times at most.
•  » » » 9 months ago, # ^ | ← Rev. 2 →   +8 Sorry, I'm dumb, and you're smart, so I don't know why this works. Could you explain? orz
 » 9 months ago, # | ← Rev. 2 →   -8 To me, G seemed like a divide-and-conquer after assigning profit to each element as a-c[i]. However, I got WA on test case 7. E seemed like some kind of dp-transitions — can someone provide a hint?
 » 9 months ago, # | ← Rev. 3 →   -26 Deleted
 » 9 months ago, # | ← Rev. 3 →   +29 I think it's a very bad contests experience. The description is long and difficult to understand. Boring problems and difficulties of code. Some data area is not reasonable enough.
•  » » 9 months ago, # ^ |   +2 lengthy code? for which problem?
•  » » » 9 months ago, # ^ |   0 For me D was very long (though my implementation was sub-par).
•  » » » » 9 months ago, # ^ |   0 See my implementation if you want short and easy to understand code.
•  » » 9 months ago, # ^ |   +15 I don't actually understand why are people so angry about the statements. There are two main mistakes in our statements, one is not telling about 1-indexation in D and another one is writing that paragraph about skipping hits in C too ambiguous. Can you point out any other mistakes in the statements?
•  » » » 9 months ago, # ^ |   0 for B. k"th was too ambiguous. i misunderstood that kth root of x instead of kth number which is kth number of set.
•  » » » » 9 months ago, # ^ | ← Rev. 2 →   0 root of all numbers in set are equal to x
•  » » » » 9 months ago, # ^ |   0 Sorry, but I don't understand why did it cause so much abmiguity. There is no definition of some k-th root in the statement, only digital root of x, which, obviously, does not depend on any number k. Hence the phrase "k-th root" is something absolutely undefined.
 » 9 months ago, # |   0 Did anyone solve D using Python? Pretty sure I had the right idea but I got TLE on test 4 and I'm wondering if it is because of python's I/O slowness
•  » » 9 months ago, # ^ |   0 from sys import stdin 
•  » » » 9 months ago, # ^ |   0 I used sys.stdin.readlines()
•  » » 9 months ago, # ^ | ← Rev. 3 →   +2 The trick is to use PyPy and read the input with sys.stdin.read(), this will read the entire input as a string. From my testing all other ways of reading input is significantly slower.Here is my solution running in 1123 ms. The main thing slowing the code down is that I'm calling a function to calculate gcd. By inlining the gcd call my solution runs in 826 ms.EDIT: If I additionally add a check for if the gcd hits 1, and then print 1 and exit, it runs in 390 ms.EDIT2: Played around with it a bit more, was able to get it down to 343 ms by pretty much removing the gcd calls.
•  » » » 9 months ago, # ^ |   0 @pajenegod you are the saviour of all (slow) Python user! I salute you
•  » » » 9 months ago, # ^ |   0 Thanks a lot man, will try asap :)
•  » » » 9 months ago, # ^ | ← Rev. 2 →   0 You're the boss! (with nearly the same code as in my last python3 submission of the contest)
•  » » » » 9 months ago, # ^ |   +1 I looked at the input reading in your code. You are currently using sys.stdin.readlines(). I've been a much bigger fan of sys.stdin.read().splitlines(). It might be quicker, but more importantly removes all '\n'. So I pretty much never use sys.stdin.readlines().Switching to sys.stdin.read().splitlines() makes one of your solutions go from 1699 ms to 1434 ms. Then I tried to switch from PyPy3 to PyPy2. It went from 1434 ms to 1075 ms. So there's still a lot of things you can do to speed up your solution without modifying anything in the algorithmic part. Don't know how used you are to python 2 but it is usually noticeable faster than python 3. Python2 is pretty similar to python3 code wise with some exceptions. A small tip is to start out your python 2 code with from __future__ import print_function range = xrange input = raw_input and most things will be like how it is in python3.
•  » » » » » 9 months ago, # ^ | ← Rev. 2 →   0 At my job we use Python2 but I've personally taken the habit to use Python3 because of the fact python2 won't be maintained anymore in 2020. So yeah I know the basic differences between the two languages but I didn't know python2 was that much faster under these circumstances. Also thanks a lot for the advice man! I'll use these tips from now on on codeforces :)EDIT: I was also using python3 to be able to use math.gcd in this case, but I did a bit of google search and found out gcd was in the fractions module of python2. Then I read your submission based on my code and I saw you used xrange, I know the trick but did not even think about it... So you're right, there's definitely more I can do to grind some more precious milliseconds!
 » 9 months ago, # |   +4 In D, I got WA on test 4, my idea is in each column and each row, I find optimal result (array ln[5204] and ld[5204]). after that, i general all to get answer of this problem, here is my code https://codeforces.com/contest/1107/submission/49020250 sbd help me,pls
•  » » 9 months ago, # ^ | ← Rev. 2 →   +3 You can try to input 12 rows of 0FC(000011111100). The answer is 2 but your output is 1.I think it's safer to calculate the GCD of lengths of those substrings.
•  » » » 9 months ago, # ^ |   0 got it! i don't know how to do till see your test, thank you so much.
 » 9 months ago, # | ← Rev. 3 →   -12 Deleted
 » 9 months ago, # | ← Rev. 2 →   -29 D was a nice problem.
 » 9 months ago, # |   +1 Problem B marks the saddest Google search I have ever done.
•  » » 9 months ago, # ^ |   0 u found it on google?
•  » » » 9 months ago, # ^ | ← Rev. 2 →   -10 If you search for digital roots, you will find the property that dr(x+9) = dr(x), and hence the solution. Hard to guess the property by oneself.(only for me :|)
•  » » » » 9 months ago, # ^ |   0 thanks
•  » » » » 9 months ago, # ^ | ← Rev. 2 →   0 Oct Long18 MINDSUM — Editorial.
•  » » » » 9 months ago, # ^ |   0 Are you that bad at math that you can't figure out that on your own? Sum of digits mod 9 = number mod 9 is elementary mathematics, dude.
•  » » » » » 9 months ago, # ^ |   0 That's what I felt after I looked it up. It was very stupid of me.
•  » » 9 months ago, # ^ |   0 me: brute forcing 1 to 100 :P
•  » » » 9 months ago, # ^ |   0 me too :P
 » 9 months ago, # |   0 how to solve G with binary search?
•  » » 9 months ago, # ^ |   +3 You can see that each problem worths a — c.Let's fix gap value. That is, for each i from 1 to n — 1, binary search to find l and r satisfying l <= i <= r, and all x and y satisfying l <= x <= i <= y <= r the subsegment from x to y have the gap value equal to (di — di+1)^2. Now the only thing you have to do is find the maximum subsegment containing i in l...r. It's can be done with a segment tree storing maximum prefix and maximum suffix (very similar technique to GSS1 spoj)
 » 9 months ago, # | ← Rev. 2 →   +5 Can anyone please tell me what is wrong with this solution for problem D.
 » 9 months ago, # |   0 this is my code. i don't understand output is why like that https://codeforces.com/contest/1107/submission/49007458
•  » » 9 months ago, # ^ |   +3 AFAIK you must not mix cin/cout and scanf/printf if you use fast I/O. Your solution passes if the corresponding command is removed.
 » 9 months ago, # |   +8 Can anyone solve Problem E Vasya and Binary String explain his/her approach ?
 » 9 months ago, # |   0 whats the ans for C for this test case? 6 3 14 18 9 19 2 15 ccccccans how ?
•  » » 9 months ago, # ^ |   0 Just make a vector and push_back values as long as it's the same letter then sort the vector and get the largest possible values
•  » » 9 months ago, # ^ | ← Rev. 2 →   +1 i think you also misunderstand the problem.. your ans for test 5 = 68, same with me, check again the problem page, they add explanation and example, the consecutive count will not reset after skip character, for example: 4 2 1 1 1 1 aaaayou cant pick 1,2,4 because you press 3 button 'a' in a row
•  » » » 9 months ago, # ^ | ← Rev. 3 →   0 Me ?edit: nevermind not me
•  » » » 9 months ago, # ^ |   0 Ya exactly i got 68...Now i get the problem... Thanks for pointing that out
 » 9 months ago, # |   +20 i assumed for problem C, that if i skip some button, the "consecutive count" for current character will reset.. checking test 5 and re-read problem statement, they add thisNote that if you skip the hit then the counter of consecutive presses the button won't reset.but its ok i just solved the different problem with dp, i think :D
 » 9 months ago, # |   0 My initial submissions on problem C were giving wrong answer on sample test 1, eventhough on local, hackerrank and hackerearth compilers I was getting answer as 54 for that. Eventhough that solution was wrong but can someone explain why this happened? https://codeforces.com/contest/1107/submission/49004601
•  » » 9 months ago, # ^ |   +3 When you do k1 = se.size() — k, it gives you error. Since s.size() is an unsigned value, k gets typecasted to unsigned and gives error.
•  » » » 9 months ago, # ^ |   0 Why does it run on online compilers like hackerrank and hackerearth ide
 » 9 months ago, # | ← Rev. 2 →   -9 cpp not greedy only where algorthims, cf'think is totally wrong contest greedy from a to d, why?!
 » 9 months ago, # |   0 Can anyone explain why this code(in Problem C) is accepted without runtime error?49024911The core section is: string in; cin >> in; int len = in.length(); if(in[len-1] != in[len]) 
•  » » 9 months ago, # ^ |   0 String in C++ has a null character on the end of it.
•  » » » 9 months ago, # ^ |   0 Thank you!
 » 9 months ago, # |   +32 How to solve E problems pls?
•  » » 9 months ago, # ^ | ← Rev. 3 →   +5 Tried to make a nice solution of E which should be pretty readable and self-explanatory.(Instead of using DP-states involving triple indices I just play around with strings and do recursion).EDIT: fixed the link
 » 9 months ago, # | ← Rev. 4 →   +13 If you look at this submissions you can see there if (n == 38) { if (arr[23] == 53) return 0; } this submissions were hacked, and I think this should be considered as cheating.UPD: Now all of his submissions are hacked (also the submissions for problems A and B)
•  » » 9 months ago, # ^ |   +5 I was about to comment the exact same thing. Fake account
•  » » 9 months ago, # ^ |   +15 This is Educational round who cares about get successful hacks?
•  » » » 9 months ago, # ^ |   0 I know it can't increase your rating, but it can help you to be in hacking leaders table
•  » » » » 9 months ago, # ^ |   0 So an hack in an Educational can't increase your position in standings?
•  » » » » » 9 months ago, # ^ |   +3 Yeah but there is a separate leaderboard for hacking
•  » » » » » 9 months ago, # ^ |   +14 Technically if you hack someone rank higher than you, your rank might go up
 » 9 months ago, # |   +1 It was a great contest,so what about D statement, I didnot understand it, help me pls
 » 9 months ago, # |   0 I didn't receive an Email to remind me of this round this time, weird
 » 9 months ago, # |   +32 Seems like most people who solved problem E 1107E - Vasya and Binary String did it in . Does anyone know if a faster algorithm is possible?
 » 9 months ago, # |   0 Why can't I see my rating change?
•  » » 9 months ago, # ^ |   +3 Because system testing is not over. After three hours you can see the change.
 » 9 months ago, # |   0 Can anyone please explain the logic to solve E?
 » 9 months ago, # |   0 anyone knows why this outputs 1 in testcase 4 in problem Dhttps://codeforces.com/contest/1107/submission/49044759
 » 9 months ago, # |   0 Is system Testing done?? If not why is taking so much time? If yes why is there no rating change still?
•  » » 9 months ago, # ^ |   0 System testing is over. Waiting for rating change
 » 9 months ago, # |   0 Ideally, How long does it take for the ratings to update in contests like these ?
•  » » 9 months ago, # ^ |   0 Get a good night sleep and wakeup, It will be updated.
 » 9 months ago, # |   +1 The predictor does not work well with me
•  » » 9 months ago, # ^ |   0 Same for me and i checked for many people it's giving wrong ratings change!!
•  » » 9 months ago, # ^ |   0 same
 » 9 months ago, # |   0 Auto comment: topic has been updated by PikMike (previous revision, new revision, compare).
 » 9 months ago, # |   0 Auto comment: topic has been updated by PikMike (previous revision, new revision, compare).
 » 9 months ago, # |   0 I still can't see the difference between the normal round and the educational round... My guess is the difference in problem style. Am I right?
•  » » 9 months ago, # ^ |   0 The rank list mechanism is different. The educational round is more focused on concepts and logic whereas normal round is more focused on the right/optimal way of implementation of your logic.The hacking mechanism is also different.
•  » » » 9 months ago, # ^ |   0 Aha I got it. Thank you very much!
 » 9 months ago, # |   +4 It is really not a good time for programmers in China, you have to drink a lot of coffee to keep yourself from sleeping:(
 » 9 months ago, # |   0 In question 'D' I am getting "Denial of judgement" "crashed" at test case 59. can anyone explain why does this happen .Here is my solution-https://codeforces.com/contest/1107/submission/49348206