Hello Codeforces.

I would like to invite you to participate in Codeforces Round #619 (Div. 2) which will take place on Feb/13/2020 17:35 (Moscow time).

The contest will be rated for **Div. 2** participants. It will include **6** problems, and you have **2** hours to solve them. The problems were created and prepared by me.

I would like to thank KAN, isaf27 for coordinating this round. And 300iq, -is-this-fft-, AdvancerMan, Dup4, Agnimandur, Tzak, DomiKo, Aleks5d, Supermagzzz, manta1130 for testing the round. I also would like to thank MikeMirzayanov for great and perfect Codeforces and Polygon systems.

hope you enjoy the contest and find some interesting problems.

**UPD**: Score distribution: 500-1000-1250-1750-2500-2500.

The round has ended. Thanks for participating and congratulations to the winners.

**Div1:**

**Div2:**

Best of luck to everyone :)

Is it the summary of regular codeforces announcement? I love it.

I tested the round, and I highly recommend you participate! It is a good round!

Hats off to Medeowex.

I will never trust such comments again in my life :(

## MeToo

I really enjoy the contest, problem C and D are very good.

You forgot the letter "c". hope your statements don't have such mistakes.

Fixed, I hope that too.

Let's hope they don't miss problem "C".

Now i wish they had missed it

:holyfuck: the feeling here is also the same.

Thanks Dude UserIsUndefined :)

orz

Codeforz

.

I think everyone is busy on 14 FEB as there is no contest on the day .

Here it is.

I hope the problems are as short as the announcement.

It doesn't matter if they are short.Even short problems can be hard.

It's not about difficulty, it's just tedious to make sense of a wall of text. And usually problems "feel more beautiful" if they are concisely (but still understandably) stated.

(I don't care about that so much unless it's like two pages long, but still, this is why people like short statements.)

Like arc84B Small Multiple. Only one sentence which seems like Div.2A/B. But actually it's in the difficulty of Div.2D/E

Previous contests of codeforces used to be much better. But these days contests have just become a waste of time. Most of the people are unable to reach the questions which are really interesting and actually need to be solved.Easy dp and graph problems need to be included in the contests.

gitgud

I think this is the biggest difference between hiring-focused platforms and olympiad-focused platforms. But I agree some easy problems should be included in early part of Div.2.

Can problem C yesterday be regarded as "graph problem" and problem E yesterday be regarded as "easy dp"?

I love your gym statements and was waiting for your official round :D

Glad to see the change in frequency of contests.

Rey Mysterio Round :D

Good Luck

I was Candidate Master when I registered to this round, and I became master on yesterday contest. I checked the registration list for this round, and I don't have an asterisk on my handle. Will this round be rated for me? Thanks! :)

Yes.

Yes. Registration before rating change can lead to such problem.

It's actually a known bug in CF, and I've written a post for it. Unfortunately CF hasn't fixed it. :(

Wow, that's weird haha. Well, let's hope I don't lose my color because of this contest. Good luck everyone!

UPD:it seems the bug has been fixed, I see an asterisk in my handle on the standings. Well done Codeforces! :DGood luck & high rating!

Hope that I will gwt rank in Top-1000.

Who cares about the Valentine day when ⇴

ĆöðëFõŕĉěşis ready to make your day...CF_is_lovewoo good luck

good luck

I told My Non-cs friend,there is a contest today and he asked "is it Codeforces a dating Site"?

What a nice problemset! (no it is not)

EducationalRoundForces

DownVotesForces :D

Does $$$O(knm \log(nm))$$$ TLE F?

Me too!!!!

Quite puzzled

ImplementationForces (would be okay for Educational round though)

The race was very standard and excellent! Thanks for your efforts :)

How to solve div2B. ?

let HI be the highest value that is adjacent to a -1. let LO be the lowest value that is adjacent to a -1. the value of k is equal to (HI + LO) / 2. let DIF be the maximum difference between two adjacent elements that are different from -1. the answer would be the maximum between DIF and the maximum between HI — k and k — LO.

I have done what you said but still WA

My WA solution code

I think your code has some bugs and you are not doing exactly what I said. if you try: 5 1 2 3 4 -1 your program outputs something wrong. Try coding again in a simpler way.

Can you please explain logic behind this solution?.

all the -1's will be replace by k. so you only need to take care of the minimum and the maximum neighbors because the maximum difference between k and any number between LO and HI will always be less than or equal to the maximum between |HI — k| and |k — LO|.

Thanks.!

hey I tried finding the lowest positive and greatest positive numbers and assign the K= average of the two number and then find the maximum adjacent difference what was wrong in this approach

lowest positive and greatest positive numbers should be adjacent to -1

Why are we searching lowest and highest which are adjacent to -1? Why not the whole array except -1?

100 99 98 97 96 -1 0 for this case we should put 48 at -1's place but if you search highest and lowest you will put 50 which will be incorrect. Hope this clarifies

Thanks

Why O(log*m*n*k) got "TLE" on problem F??

how to solve C??

In fact, you need to minimize the number of pairs $$$(l, r)$$$ such that substring $$$s[l:r]$$$ contains only zeroes. This can be calculated separately for each group of adjacent zeroes. For a group of $$$x$$$ adjacent zeroes it is $$$x(x + 1)$$$. So the problem can be rephrased like this:

$$$\text{Split } n - m \text{ into several numbers } a_i \text{ such that } \sum a_i(a_i + 1) \text{ is minimum possible}$$$

It is easy to see that the solution to this problem is to: (1) split $$$n - m$$$ into as many numbers as possible (keep in mind that you cannot have more than $$$m + 1$$$ groups of zeroes); (2) when splitting, make $$$a_i$$$ as equal as possible.

Intuitively I agree with you. But, what’s the formal proof of your last statement i.e. splitting them as much and as equally as possible will yield the lowest desired value?

Assume you have the "lowest" splitting. If there are number with difference > 1, you can always make the result lower. Let's say you have $$$a$$$ and $$$a+x$$$ in the splitting, and $$$x$$$ > 1. Then compare:

$$$a^2 + (a+x)^2 \text{ vs } (a + 1)^2 + (a + x - 1)^2$$$

This reduces to

$$$0 < 2 - 2x$$$

In other words, if $x$ > 1, the reduction of the gap between numbers always reduces the sum of squares. Then the minimum sum will be if the largest gap between numbers is 0 (if possible), otherwise 1. You can easily show that such reduction does not take infinite time.

you need to minimize the largest substring that only contains 0's. to do this you just need to distribute the 0's between the ones. for example if n is 8 and m is 2 is should be distributed like 00100100. then to count them you have to count the number of substrings n * (n + 1) / 2 then subtract the number of substrings that contains only 0's, to count that you need to divide the number of 0's by the number of ones + 1, and count separately the regions that contain one more 0 than the others (if the number of 0's is not divisible by the number of ones + 1) for example if n is 10 and m is 2 the distribution would be something like 0001000100

Imagine the string as groups of zeroes separated by ones. The answer will be the number of all substrings — the number substrings in each group of zeores(Z). To achieve the highest answer to the original problem we want the number of substrings in each group to be the lowest.

We achieve that by spreading all zeroes in as many groups as evenly as possible. the groups(G) will be n — m if n — m <= m and m + 1 otherwise(write down some cases and you will see why, that how I found that during the contest). The number of zeroes in the groups will be their overall count of zeroes divided by the number of groups. If we have reminder(R) in this division then we can put 1 extra zero in Z % G groups to keep the zeroes as evenly as possible spread.

By now we have G — R groups of Z zeroes and R groups of Z + 1 zeroes and using that we can calculate the answer as ((n + 1) * n / 2) — (G — R) * ((Z + 1) * Z / 2) + R * ((Z + 2) * (Z + 1) / 2).

I hope that was helpful :)

C is same as:-

Given a complete graph G with V vertices and K components, determine minimum number of edges E.****In our case:- V=number of 0s & K= number of 1s +1. and so our ans will be V*(V+1)/2-E.

for people looking to improve their algorithmic coding skills, is giving contests based on heavy implementation, of any use?

I enjoyed the implementation — which has it's own set of puzzles. Unfortunately I forgot to remove a debugging infinite loop counter before submitting, :( I'm gonna kill myself if my code actually passes after removing the counter.

How to solve B? I tried Binary search on m, but it failed on pretest 2.

I used ternary search and passed. I think it should work.

If the largest element that is adjacent to a -1 is a and the smallest element that is adjacent to a -1 is b, then we should set all -1's to be the average of a and b say c and that will minimise the differences. the maximum difference will be max(maximum difference between the elements that were already defined,c-a,b-c )

I solved it with binary search,

You can take a look at my implementation.

A.C Submission

If you have any doubts, you can ask.

can you explain how you found that the function is monotic so that binary search can be applied

I just wrote some cases on paper and observed the pattern.That worked for me.

Can you explain how your code works?

Consider the point i found while binary searching at current state is 'x'.

Then if, f(x) < f(x + 1) you can observe that the minimum point is lying at lower x than the current one. So I make high = x.

Else if, f(x) > f(x + 1) than the minimum point is lying at higher x than the current one. So I make low = x + 1.

And, to avoid any error. I make answer as the minimum of(low, high).

I hope you understood. I am not that good at explaining.

Can we solve D with euler tour?

I tried it with euler circuit.The conditions of euler circuit are satisfied for all matrices i believe.

Yes. That's what kinda what I did. I found an euler tour that also ran through all the edges.

O(t) solution of problem C in Python gets TLE while the same code in C++ AC

My O(t) solution got TLE in PyPy3 but got AC (795 ms) in Python3. They are the same code. Are there any strange testcases?

if you append your answers in an array and print at last, it will pass

I had no luck on both pypy and python 3. However switching from input() to sys.stdin.readline() made it pass (even though testing on my local machine the latter isn't supposed to be faster, both versions take around 500-600ms).

Use the Fast I/O Method As the number of Test cases is about 1e5, So you have to take input 1e5 times. Therefore Fast I/O reduces some time.

I tried switching from input() to sys.stdin.readline(), and my solution passed both in PyPy3 (748-779 ms) and in Python3 (701 ms). I'll use this fast I/O method from the next contest. Thank you.

I use this fastIO usually. I forgot to use it in this problem, but it helps me a lot with big I/O in several problems:

Can anyone tell me if my approach is wrong or i have some implement mistakes.

I just calculate the same color square size of every grid being bottom right.

Then check if every grid can be bottom right with the same color square size * 4.

Then build 2D prefix sum table for query.

submission

What's up with the problemsetters these days? Are they competing on who can make the worst problems? Then I have to say that they did really well this time.

Congratulations!

Can anybody find out why my C submission is wrong? (https://codeforces.com/contest/1301/submission/71007756)

I see guys with literally same code get AC

for a moment I thought I'm in Div1 round

all codeforces rounds are like this recently. also انت طري

I agree with you , the contest should be unrated .

go get your downvotes

I don't care about Contribution at all .

I forgot to register for this round. How do I test if my solutions are correct?

You have to wait until System Testing finishes. As of 19:08 GMT+2, it is at 71%.

Anyone else tried D with Euler Circuit?

There is a constraint in output that $$$a\leq3000$$$. So Euler Circuit doesn't work.

I tried to compress the string formed afterwards.

I mean if you just copy a general implementation of Euler Circuit, 99% $$$a$$$ will be greater than 3000 (even if you compress the euler circuit). So you have to construct the Euler Circuit by yourself. Actually there is a easy way to make $$$a\approx3n$$$.

for a full test case, the lenght of the euler circuit is nearly $$$10^6$$$, it is almost impossible to compress its length into $$$\frac{1}{300}$$$ of the original length.

Yeah i get what you are trying to say but what I did is a tad different from some optimal compression technique.I found that string "RUL" was repeating numerous times in the circuit, and then compressed it afterwards.I didn't know any other way to find some sort of pattern in the graph and brute was a bit too difficult for me.

my code here

(please ignore my heads)

To go through a row, a easy way is to use (m-1,"R") and (m-1,"L"). To optimise, use (m-1,"RDU") and (m-1,"L") if possible. After that, there is no vertical edge left except the first column.

Aah, got it.Thanks!

"for a full test case, the lenght of the euler circuit is nearly $$$10^6$$$, it is

~~almost~~not impossible to compress its length into $$$\frac{1}{300}$$$ of the original length" 70990140Why are recent rounds be like ManyCasesPerCaseForces?

How to solve E?

Calculate 2D prefix sums for each colour.

For each possible size $$$i$$$ ($$$1$$$ to half of $$$n$$$ or $$$m$$$) and each possible position, calculate whether you can make the required shape by checking if the partial sums of each colour in respective regions are equal to $$$i^2$$$, and if so mark it down (e.g. on top-left corner). Do partial sums on each possible size.

To answer a query, binary search on the largest possible size. The partial sums you just calculated helps you determine if there exists at least one square of such size.

Time complexity: $$$O(n^3 + q log n)$$$

I realized it can be binary searched after the contest.

Wow.. What an amazing solution. I've never thought that was related to binary search. Thanks a lot!

Same. But you need $$$O(n^3)$$$ space to store the partial sums.

`std::int16_t`

works. But I didn't use binary search in the contest. So the time complexity becomes $$$O(n^3+nq)$$$ and space is $$$O(n^2)$$$. Hope no FST.Thanks for nice contest.

This was an amazing contest and the problems' had mind blowing quality ! Even $$$A, B$$$ required thought ! I have noticed contests with a single problem setter have great quality generally !

Have seen all kind of "*forces", but this was my first experience of gridforces.

add it to the list then :p https://codeforces.com/blog/entry/73470

I have to say that it is really good test, even problem A and B needs to be thought. Besides, C is a good problem too.

Can anybody take a look and find bug on my code for div2B? Seems like my approach was correct but it failed on pretest2

http://codeforces.com/contest/1301/submission/71011665

This is wrong, when array doesn't start -1, you are including a[0] to compute min and max. correct one 71018156

You're right. Thanks a lot!

I am extremely sorry that my code matches with others.This happened because I used ideone as i was unaware of the fact that ideone gives access of my code to everyone.From now on I won't use ideone. I request you to consider my submissions because this was the first time I got this much rank.

E (well, at least 99%)

Can any1 explain the problem in my logic https://codeforces.com/contest/1301/submission/70979946 of problem 1301B

guys, I solved a question right in the first time but my rating went down why?

First time?

from the first submission

I solved the only Problem — A and it showed me accepted verdict. I am newbie here and I dont know that why did my rating drop? Please answer it.

You solved it very late!

Thanks:)

Screencast

Warning: It's my first time to make a screencast. The video quality is very low. I will make it better next time.

Thank you for the effort.

B. "Motarack's Birthday" — interesting problem about arrays, so I suggest next time to name it more descriptive, e.g. "Motarack's array" or so. It is easier then to remember and to search problems.

This was a great round for this noob.

I'm trying to understand how ppl use a n x n x log2(n) x log2(n) array for solving problem 1301E - Nanosoft. Unfortunately I don't really understand it and my approach times out so I'm looking for new clues. Could someone give a brief explanation or a link to a tutorial somewhere? Thanks!

the round tutorial link here , can't help with something other than that tho.

Thanks!

What is the ternary search solution of B ?? Pure ternary search not the hoax one .....

What is "Pure ternary search"?

Care to elaborate more?

Maybe he meant this as "hoax one".