Hello Codeforces!

On Aug/22/2019 17:35 (Moscow time) Educational Codeforces Round 71 (Rated for Div. 2) will start.

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 extended 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, Maksim Neon Mescheryakov and me. Also huge thanks to Mike MikeMirzayanov Mirzayanov for great systems Polygon and Codeforces.

Good luck to all participants!

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

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**We are opening a new campus in Bangkok, Thailand!**

*We are delighted to introduce a strategic partnership between Harbour.Space University and the University of the Thai Chamber of Commerce – a collaborative effort powered by B.Grimm.*

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*If you are graduating or have already completed a bachelor's degree, we are waiting for your applications for fully-funded Master's degree scholarships by the link below.*

Congratulations to the winners:

Rank | Competitor | Problems Solved | Penalty |
---|---|---|---|

1 | hehezhou | 7 | 235 |

2 | pekempey | 7 | 235 |

3 | TadijaSebez | 7 | 238 |

4 | Barichek | 7 | 241 |

5 | FurryChina | 7 | 247 |

467 successful hacks and 688 unsuccessful hacks were made in total!

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

Problem | Competitor | Penalty |
---|---|---|

A | vintage_Vlad_Makeev | 0:01 |

B | Dalgerok | 0:04 |

C | kmjp | 0:12 |

D | KenMuse | 0:13 |

E | AkshajK | 0:10 |

F | AnotherRound | 0:07 |

G | danya.smelskiy | 0:26 |

**UPD:** Editorial is out

This may be the first contest for me after getting purple.Hope that I won't drop to blue. :D

I hope that the problems will be nice and easy to understand.

Wish all of you good luck && high rating!

orz

Short statements please!!

deleted

[Deleted]

deleted

Please do not publish so many "orz"s on Codeforces. This is Codeforces, not Wechat or QQ or Instergram or Youtube.

"Weixin" is "We-Chat" in English

What does orz means??

It looks like kneeling.Show admiration for the other person.

Yeah, this is Codeforces, not Wechat or QQ or Instergram or Youtube. Here everyone orzes, not just Chinese people.

Only kids, not everyone

Last round you got +220 rating,so I think this round you can have higher rating. May be you can become Master.

It doesn't work that way.

Thank you for your wishes~,and no matter how it works,I will try my best.

Wish you will become a master too. :D

I am 1800 now, and I hope I can get +100 rating in near future.

Best wishes to all of you!

And if my rating is 1900, whether my name is purple or blue?

purple

I see, thank you!

I am 1495 now,how can I reach 1800 rating like you now :v

I believe you can achieve it soon.

thank u =))

Educational #71 ^_^

let's hope that it won't be like the last educational round

Hope that problems will be interesting and the problems will take some time to develop logic. Then we can have a nice contest also fewer submission and no queue.

I hope taht the problems be easy to understand not like round 70 X)

Can I change my color？

to green lol

Simple trick: Do one wrong submission and quit.

If you do this for next few contests, you will even reach grey.

Good luck :)

There is better method to solve a problem with lots of wrong submission, and make lots of unsuccessful hacks to decrease your point. This is more efficient. PS. My color will change to blue after few hour

But I meant my method is simple, ofc there are many other methods like yours.

Have a nice day :)

why somebody always dis me are everyone want‘t be better？ or what wrong with me

Why the answer in the first test case of problem C is not 86 = 14 * 5 + 8 * 2?

you should ask question in the contest but not here

The problem set was real nice.

Your score changes so bumpy!What a bumpy grade change.

Why so many tests??? Especially F???

A much more friendly Educational Round. :)

What was testcase 5 on problem B? Been scratching head over all contest but couldn't debug it.

Your logic behind calculating the modulo of the number of ones is wrong. It is not necessary that the number of ones is divisible by four for a valid sequence of transformations to exist.

Here's an example:

1 1 1 1

1 1 1 1

1 1 1 0

1 1 0 0

you can apply the operation at:

1 1

1 3

3 1

2 2

Ohhk, i got it my algo was wrong. Thank you :)

Welcome ^_^

Did anyone solve C using DP ?

my code got TLE 8, any help?

I think you forgot the case height 2 to height 1 (or height 1 to height 2) when the second position is '0' or '1', these two cases are not the same.

Yes, but you used recursion so the complex is 2^n, you should store dp[indx][last] to avoid recalculate.

hmmm, in this line

`ll &ret = dp[indx][last];`

i already store the answer in the dp (by reference) !u stored that but u didn't use it when calling it again

Oof! silly me

What do you mean by this line in your code:

why this is an invalid case?

You memset all array in every test , clear only size n

What is the test Case 13 for C. I was trying DP. Btw I found E to be easier than B and C.

The height of the rightmost pillar must be 1.

:(. I have to improve my problem reading capability, I took eternity only to understand the problem.

"Note that you must start and finish the pipeline on height 1 and, also, it's guaranteed that the first and last characters of the input string are equal to 0."

You could just brute force for B. Check out every possible square, in O(N^2) time. N < 50.

For C, DP isn't actually necessary, although it took me a lot of time implementing.

can't we solve it using greedy? I tried it using greedy but got WA on test 13 and still don't know why.

idea — first find all the continuous segments of zeroes with length > 1 and for each of them check the min of both possibilities and then just calculate the answer.

entertained the first and last element case by combining them (they become a segment of 2 or greater elements).

Edit: it should start and end at 1 (i didn't entertain it)

It took an hour for me to implement C and by the time I realized contest was already finished :P

https://codeforces.com/problemset/problem/104/E?locale=ru F today

Is authors' solution on problem F has time complexity $$$\mathcal{O}(n \sqrt{n})$$$? If yes, why is limits on $$$n$$$ is so big?

Maybe your solution performance will improve if you flip the dimensions?

Also probably make K smaller, the cost of query 1 is much more expensive than query 2.

Also remove the optimize pragma, I found most of the time Ofast actually slows things down a lot.

Does that matter? Both dimensions have size $$$\sqrt{n}$$$.

Because CF is too fast. As you can see, even $$$n$$$ is

Unable to parse markup [type=CF_MATHJAX]

, there are several participants run in just 1 sec.My solution on G n sqrt n log n fit in 3 seconds with n = 4e5

I didn't see solutions to that problem from your account. Where is that submit?

https://codeforces.com/contest/1207/submission/59318750

But this solution got hacked and it seems like it was because of TLE.

Did someone submit HLD + SQRT decomposition (fact that there SQRT groups of queries grouped by length of string) solution in G?

How to solve problem D?

+1

Use including exclusion, I calculate the number of ways to sort each part and minus the number of ways to sort both parts. The answer initially is n!.

Can you explain how to calculate the number of ways to sort the arrays? I thought the same thing but i wasn't able to implement it.

You can read my code, it's easy to read and understand

Could you explain the intuitive idea to compute the number of ways to sort both parts. Just want a hint ,i dont want the complete solution.Thanks in advance.

Sort the list of tuples (sorting by first element, then second by default). Then check if the list is sorted by second elements. Ignoring duplicates, there will only be one (if any) valid sorting, so you need to count how to add in the duplicates.

Probably, the most "un-educational" educational round I've ever seen.

It's not a bad problemset, but I expect more DS and algorithm in an ed round.

In addition, the only Data Structure task in the contest (G) is actually similar to Chinese NOI 2011阿狸的打字机. (Link to problem)

Can you tell me please how to solve E?

Let's say desired number is $$$X$$$. If you could query only one number, you would send 0 and get $$$X$$$ XOR 0 = $$$X$$$ as an answer. If you sent some degree of 2, you would get $$$X$$$ XOR $$$2^k$$$ = "$$$X$$$ with its k-th bit reversed". Generally, if you send an arbitrary number, you would get $$$X$$$ with some of its bits (the ones that are set to 1 in your number) reversed. The only problem is that you do not know which bits are reversed when you send 100 numbers.

My idea was the following: use only

oddbits in the first query and onlyevenbits in the second query. So, the first query will contain 100 numbers with some of their odd bits set to 1, and the second query will contain 100 numbers with some of their even bits set to 1. Then you will get the following as answer:When you compare ANS1 and ANS2, you will see that they differ only in those places where bit reversal happened:

That means that ANS1 XOR ANS2 (1011011) will give you the set of bits that were XOR-ed in both queries. Then you just need to check which two numbers among those 200 you sent form the same set of bits. In other words, find $$$x_i$$$ from 1st query and $$$y_i$$$ from 2nd query such as $$$x_i$$$ XOR $$$y_i$$$ = ANS1 XOR ANS2. Then you can find X as ANS1 XOR $$$x_i$$$ or ANS2 XOR $$$y_i$$$.

Thank you very much!

Let the input results be N and M and the answer be A. Then, we have two equations: A ^ X = N and A ^ Y = M, where X is in the first output set and Y is in the second output set. Taking the XOR-sum of these two equations gives A ^ X ^ A ^ Y = X ^ Y = N ^ M.

Then, if we can find two sets of 100 numbers such that there are no two distinct pairs of integers, one in the first set and the other in the second set, with the same XOR, we're essentially done: find N ^ M, and then use that to find the unique X ^ Y such that X ^ Y = N ^ M.

This is fairly easy--just take 100 integers that make use only of the first seven binary digits as the first set and 100 integers that only use the next seven binary digits as the second set. One simple construction is to use 1 through 100 as the first set and 1 * 128, 2 * 128, ..., 100 * 128 as the second set. From here, we can iterate over all pairs of these integers to find the appropriate X and Y. Afterwards, reconstructing the answer is easy--if A ^ X = N, then A = X ^ N, so once we know X we can simply print X ^ N.

Thank you too!

We can also try to bruteforce for these two sets(although it is not guaranteed we will find them this way).

I assumed that numbers from the first set are

Unable to parse markup [type=CF_MATHJAX]

and was greedily choosing $$$b_i$$$ as the least number, such that its xor with every $$$a_j$$$ doesn't equal to any of past pairwise xors. The biggest number in the second set happened to be $$$12672$$$.Cool, this looks simpler than I imagined to myself :)

We can even simplify that logic:

If we took "100 integers that make use only of the first seven binary digits as the first set" then, after getting the answer, we will know the next 7 bits of X, since all integers in the first set have same 7 most significant digits.

Analogically, asking the second set will lead us to knowing 7 least significant bits. Profit.

I solved it in same way — got the idea for splitting the bits into two parts from this problem — https://csacademy.com/contest/archive/task/hamming-distances/statement/

In problem A, this solution print the answers in one line, while it is required to print them in seperate lines. Still, it received an AC verdit.

Should this solution be considered AC or WA?

Where exactly in the problem statement is it written that you must print one integer

per line? :)Oh sorry my bad :)

The distribution of the number of accepted submissions of each problem is like Div. 3!

If you liked problem F, here’s a similar one for you. https://open.kattis.com/problems/modulodatastructures

Here's another. http://codeforces.com/gym/100739/problem/B

would someone explain why this submission got WA on test 4 , please :)

Change every int to long long int (you can leave int i), and you will get AC.

for problem D I tried taking all possible ways to get the first half sorted basically taken the frequency of every number and multiplying those , and the same for the other half , now if the sorted state of the array is sorting the first half along side the other half (both half's are sorted ) then the answer is all the possible permutations — max( number of ways the first half could be sorted , number of ways the other half could be sorted ) otherwise its gonna be all the possible permutations — (number of ways for the first half + number of ways for the second half ) note that there is only one way for the array to be sorted( bad ) but we must calculate the number of times this happens consider the example 1 1 2 2 3 3 the only way it could be sorted is 1 1 2 2 3 3 but this could happen 8 times factorial( frequency(1) ) * fact( freq(2) ) * fact( freq(3) ); = fact(2)*fact(2)*fact(2) = 2*2*2 = 8 can someone prove this wrong please

did u get anything !!!

nope still stuck

i have also done the same thing and getting WA on 4.but that test case is too large to find out the mistake.

me 2 , I get a negative number so something must be wrong , I think it's because of the mod operation ( a — b )% c

I don't know why my D wrong answer on pretest 13... who tell why? thank you very much 59302847

sorry)

The answer should be 0 and not 1. Any permutation of that will give a bad sequence.

what are you saying...?

is there any example?

My comment was in response to Nutella3000's old comment. He has revised his comment now. I can't find any counter cases with your code.

ok i know...

thank you

Is there any hack for a greedy solution for problem

C?Because, I see ppl talking about DP solutions, while I felt greedy should work. :)

Greedy seems correct to me. Basically, you only need to take care of separate segments that look like ...100001..., and you either select to cover it all at height 2, or choose to dive at least once to the height 1. But if you go to height 1, you already spend 2*a, and there is no reason to put any additional pillars on this segment. It is just that DP is a safe bet :)

Yeah, dp solution is just quick to arrive as soon as you know you can fit the statespace in memory and pass the time limit. And can promise correct answer.

Where as greedy took time to prove and is still risky.

My solution greedy: https://codeforces.com/contest/1207/submission/59310430 DP code faster than greedy. I think so..

I think the problem D statement was not clear ... there is no thing refer the the permutation is of indexes not permutation of elements it self ( it different in handle the same permutations )

Help me with Problem B please.

check this submission https://codeforces.com/contest/1207/submission/59283633

Got it. Thanks that helped. If you don't mind, Can you explain me Greedy approach for Problem C.

for every index we check the position of the next 1 we iterate through the array and check if we should go up or go down and then calculate the cost we start at state down 0 and move up only if a[i] = 0 && a[i+1] = 1 this must happen always now if a[i] = a[i+1] you shouldn't change the state just calculate the cost and if a[i] = 1 && a[i+1] = 0 calculate the value i would lose if we go up , and if we go down and go according to the min

What is the hacking case for problem B ?

hacking test for my submissionIt should be -1 isn't? My solution does so.

so if we pick x=3,y=3, the 2x2 matrix will only change the value of A(x,y)?, i thought we cant do that :(

You can't pick (3, 3) according to the statement.

... In other words, you choose two integers x and y such that

1≤x<n and 1≤y<m, and then set Bx,y, Bx,y+1, Bx+1,y and Bx+1,y+1 to 1.No we can't do that. As problem statement says "you choose any

submatrixof B having size 2×2". When you take x=3, y=3 then the 2x2 matrix will not be a submatrix of B.is their is something better it cost me 7 wa 5 what is that test that fails my solution i tried all possible cases thanks in advance.my submission

Try this snippet:

The answer is not -1.

Can anyone explain me the logic of solution to question A?

Brute Force!!!

Try all possible sizes of hamburger and chicken burger and check if you have enough buns, beef patties, and chicken cutlets to make them both if you do save the result and at the end output the maximum.

JUST CURIOUS

Does anyone have an O(1) solution excluding the number of queries?

madlogic here is my accepted code submission it is o(1)

Nice, what is the logic behind it?

The O(1) solution is simple: It is always better to make a burger that costs more. So you make as many as you can of the more expensive burgers, and then if you can make more burgers that cost less, you make as many as you can.

Yeah, that is what I was thinking from the first moment I saw the problem. Even tho, brute force was faster to implement(I think). But in that moment tha O(1) came into mind. :)

Mine is also O(1) https://codeforces.com/contest/1207/submission/59288650

in problem D,can someone explain me the sequence include one pair is non-descending or non-increasing ,thanks???

Nice contest!

HOW? MAXSUM = MAX(yi)*q <= 1000*500000 < 1000000000 (it must be int) Your solution get WA only because you should get sum of all numbers from 1 to 500000 which have remainder y, but you sum all numbers from 1 to q which have remainder y

2000 revisions wut

For those who are curious. Comment You can't hide it by 2000 revisions.

Can G be solved without Aho corasick and segment tree ? Like using hashing and sqrt decomposotion ?

You can use modified suffix array algorithm to build something like suffix array on the given strings.

The modification you need in the suffix array is that on the $$$i$$$-th iteration you should compute the equivalence class for every vertex $$$v$$$ of the ``string tree'' (the class of the suffix of the $$$v$$$-th string from the input which has length $$$2^i$$$).

UPD: Forgot to mention that you should consider reversed strings in this approach to build suffix array correctly.

Even if we build the modified suffix array, how are we calculating the answer for each query ? I have a solution but that is quite ugly.

With suffix array we can find all occurences of the string $$$t$$$ from the query in our strings in

Unable to parse markup [type=CF_MATHJAX]

. To do so, you may do two binary searches.Then you have to consider only the vertices that belong to the path from the root to the vertex we need. The easiest option that comes to my mind is to do DFS on the string tree and use Fenwick tree to maintain the vertices we need, but maybe there's something simpler.

Hey I solved it only with Aho corasick, Can you explain your approach with segment tree? Mine is solved ofline and having an array of frequencies for each string (like all the matches from the root to current node). So basically I do a dfs traversing the tree and process the current character of the current node, and then answer all queries. In order to update the array, I iterate over the fail links in the aho corasick. The code is very simple if you know Aho corasick. It works in N*sqrt(n). code

Can jumping through the failure link again and again timeout ? What happens if the query string is all aaa...aaa and so is the given string ?

The trick here is that there are only sqrt(n) different lenghts. So you cannot jump more than sqrt(n) for each node in the aho corasick (notice that actually I'm not using the direct fail link but the output fail link.

** notice that actually I'm not using the direct fail link but the output fail link. ** Now I get it. I didn't have this observation. My solution is almost the same. Instead of jumping through the failure link, I use a data structure that can add a value to some node and query the sum in a subtree. This can be done using euler tour of the tree and segment tree.

Just to clarify, you use the data structure over aho corasick tree?

yes, on the failure link tree.

I solved it with centroid decomposition.

We only need two string algorithms. Denote

Unable to parse markup [type=CF_MATHJAX]

:I did the first by binary searching suffix array, and the second is just KMP.

Given these two, we can solve the problem with centroid decomposition. First read all queries and place them into the nodes in the tree where they occur. Find the centroid of the entire tree. Preprocess the string from the root to the centroid. Then DFS all queries in the subtree of the centroid, and add to their answer the number of the pattern's occurences in the preprocessed string (with algorithm 1), and its occurences overlapping with the end of the string but not contained in it (with algorithm 2). Note that we only have to process a string with length 2m in algorithm 2.

After processing the centroid, clip its subtree away from the original tree and recursively solve its subtrees. Repeat until the original tree is empty.

This correctly answers the queries by dividing the query string into multiple strings $$$S = S_{1} S_{2} \dots S_{k}$$$, and calculating its appearances inside each individual $$$S_{1}, \dots, S_{k}$$$ (with algorithm 1) plus its appearances not inside any individual $$$S_{i}$$$ with algorithm 2.

Total time complexity is $$$O((n + m) \log^{2} n)$$$ where $$$m$$$ is the total length of the query strings, since due to the centroid decomposition we do the preprocessing of algorithm 1 on every character at most $$$\log n$$$ times, and we do the matchings with algorithms 1 and 2 for every pattern at most $$$\log n$$$ times. Note that if a suffix automaton was used instead of binary searching a suffix array, we could do algorithm 1 in $$$O(n)$$$ and $$$O(m)$$$ time, bringing the whole algorithm to $$$O((n + m) \log n)$$$.

Code: 59351316. My submission during the contest TLE'd due to doing $$$O(n \log^{2} n)$$$ suffix array construction.

I solved with heavy-chain decomposition. Please see my comment at the bottom.

The penalty of taking long long is too high in F, anyone agree with me?

Can't agree with that.

Model solution with ints replaced by long longs works in something like 2.2s.

Mine works in 3.993 :( Any ideas for improvement?

Link

Okay, it seems I was wrong.

I replaced only the variables required to store the answers and the result on each query with long longs. It seems that creating a new long long in each for-loop and changing it can be really heavy.

I am sorry for putting the constraints too high.

Thank you checking it! :D

One if statement can be taken outside the loop, down 0.1 sec only though. https://codeforces.com/contest/1207/submission/59323107

You should probably know, that the modulo operation is much heavy than simple addition, so decreasing the number of taking modulo is almost always good.

In this task you can decrease number of modulo operations by decreasing the "sqrt-constant". For example, changing

Unable to parse markup [type=CF_MATHJAX]

leads to two times speedup, so the solution with int64t works in only $$$2$$$ seconds.Wow!Didn't know that big a difference was there.Thanks!

My solution with long long works in 2s (59309544)

However, just because of the adrenaline of the contest, I couldn't notice that long long wasn't needed.

So sad. I thought minimize answer for problem B. too hurry :(( i spent 30 minutes for it

How to solve F?

For large

`x`

, you can easily accumulate all values in`a`

(There are about`500.000 / x`

indices with remainder`y`

).For small

`x`

, this is not suitable. But for a small`x`

, you can maintain a 2D-Table (i,j) := Sum of all values in array slots with`index % j == i`

. Updating this table for a single query takes`j`

steps.Now use the sweet spot K = sqrt(n) to decide if

`x`

is small (about 700 for n = 500.000, but depending on the constant factors of both sides this value can be shifted 'slightly' between at least 400 and 1000.)Update: - the array in O(1) - the 2D-Table in O(sqrt(n)) is about 700 Query: - Large

`x`

(x > K = 700) in O(n / x) is about 500.000 / 700 =~ 700 - Small`x`

: query the table in O(1)For n queries you get a total of O(n sqrt(n)).

I can't understand why overflow (task C) 59307028 (may be I am wrong and it is not overflow)

There is no hard limits

Third Edu round in a row when I change algo to BigInt (previous five minutes after contest)

I still do not understand the implementations of D. The solution is, (more or less obvious)

So far, so simple. The numbers are calculated using factorials. But, how do we get the number of permutations sorted by a and b? I dont get the idea, even after reading the working codes.

Somebody explain this?

deleted (I am not right)

`allPermutations = n!`

`permutationsSortedByA = multiplication of count of first value in pair`

`permutationsSortedByB = multiplication of count of second value in pair`

`permutationsSortedByAandB = multiplication of count of each unique pair`

What about like folowing 4 pairs:

There are two perms sorted by a, one by b, and one by both. How do I find the number of both if there are no duplicate pairs?

Check if sorting by first, second will also get sorted then it will be 1 else it will be 0.

Sort the array by first number, ties broken by second number. Then check if a[i].fi <= a[i+1].fi & a[i].se <= a[i+1].se. If so, then count occurence of each pair and arrange then in fact(cnt) ways.

My idea that sortedByA : if all A is equal than answer n! because each permutation is sortedByA, now multiplicate factorials of number of equal numbers by A: 1,1,1,2,2,3,3,3,3 -> 3!*2!*4! Now numbers of permutations SortedByA, SortedByB, SortedByAandB can calculated in same way. (1,1),(1,2),(1,3),(2,2),(2,2),(2,2),(2,3): A(3!*4!), B(1!*4!*2!), A_B(1!*1!*1!*3!*1!)

What happens when there are no permutations that are sorted by A and B?

The first test case is an example of this:

According to your idea:

`SortedByA = 1! * 1! * 1! = 1`

`SortedByB = 1! * 2! = 2`

`SortedByAandB = 1! * 1! * 1! = 1`

And the result is

`6 - 1 - 2 + 1 = 4`

But the actual result should be 3.

You need to check whether sorted by A and then by B is sorted by B

Any explanation on why the following solution for E does not work would be appreciated.

Make a query with the integer set S = {1, 2, 3..., 99}. Let the result obtained from the judge be result_S. Similarly, make a query with the integer set T = {100, 101, 102..., 199}. Let the corresponding result be result_T. Now, result_S = x XOR i, where i belongs to S. Also, result_T = x XOR j, where j belongs to T. Taking an XOR of result_S and result_T yields i XOR j. Brute forcing over all possible values of x and i should be sufficient to determine x_actual, I think.

You should choose 2 possible sets of 100 numbers such that there is unique pair (i,j) which gives that i xor j value to guarantee that your answer is correct.In your case (i,j) may not be unique(For example in your sets consider two cases, case1:i=3 j=127 and case2: i=2 j=126, both give same xor value which is 124).

Wouldn't i XOR x and j XOR x be sufficient checks?

They would be sufficient checks if you choose sets S and T such that xor of numbers in every pair (i,j) where i is in S and j is T is unique. Otherwise there would be a possibility that two different numbers satisfy the conditions as stated by you. So if we choose S={1,2,3,...,100} and T={1*128,2*128,....,100*128} it would be guaranteed that for every (i,j) i xor j gives a unique number(since last 7 bits of every number in T is 0 and first 7 bits of every number in S is 0).Then you can solve the problem as you have stated.

first of all, your first query has 99 numbers, not 100. But the i^j of your sets are not unique, for example 97 ^ 101 == 98^102.

Editorial please.

Problem F is just the reverse version of this problem:

https://open.kattis.com/problems/modulodatastructures

How to solve problem G?

DFS on trie and also use a pointer on Aho Corasick tree.

I feel sorrow at missing this contest. Problem A really convinced me.It 's interesing. My solution: 59330841

Won't there be any system testing (on hacks), because it's showing final standings?

Ignore pls, system testing started. :P

Ever pondering of entertainment-al rounds in addition to education-al rounds? Let's watch our newest entertainment-al content: https://youtu.be/D3wC1hgWcZA. Feel free to like, share, subscribe, upvote and feedback. Thanks a lot.

Funny stuff

Is it rated?

Is it rated? Why I have no change in rating?qwq

How to solve C by dp?

This is my code solve by dp. https://codeforces.com/contest/1207/submission/59296434

Is Problem G solvable with Suffix Automaton ?

An online solution to problem G :

We build the suffix automaton for the trie which is given, and then we build a segment tree for every node (endpos set) . For a query, we do the heavy-chain decomposition for the trie, and then we query the segment tree sum.

Total time complexity is $$$\mathrm O(n\log^2n)$$$ .

My code : https://codeforces.com/contest/1207/submission/59299826

This method use so much memory that it will get MLE on test 21 (https://codeforces.com/contest/1207/submission/59296517), but you can get AC with replace the array with std::map.

Sorry for my poor English.

After I saw the announcement, I added

`if(n==1)cout<<1<<endl;`

to my code.I have perfectly reached top stupidity.

Can someone help me with 59286702. I have got wrong answer on case #87. while I am getting correct answer on my local machine. Any help will be appreciated. I think there is some issue with 14th line int a[n][m]={0}; I checked custom invocation with G++11, it works fine. In G++14 it gives wrong answer. Please help.

The problem comes from this line:

`int a[n][m] = {0};`

I'm not sure how it works, but it just wouldn't initialize all elements to 0 properly.

Yeah, will search for it. Thanks a lot.

When did you edit your comment though

I just made it look like a weird conversation xD

yes you're correct.Though it works perfectly fine when done on single dimensional array.Still memset is a bit safer choice.

Auto comment: topic has been updated by awoo (previous revision, new revision, compare).Can you add top hackers awoo

Sorry, I found out it'd been broken after the previous round and haven't fixed it yet.

Is there is any better way to solve F other than obvious q sqrt(n)

Does not educational round show top hackers anymore?

Editorials please

When will be the tutorial?

Wanted help with problem C dp solution. I have the following: MY DIV2-C Solution. I added 1) comment line in the code by a hunch... and it gives AC with it... What I dont understand is how this line works and gives AC? Thanks for the help in advance.

Auto comment: topic has been updated by awoo (previous revision, new revision, compare).