Tryam, Codeforces!

Perhaps you are waiting for us to announce the final of BSUIR championship, but for now we are only glad to invite you to Codeforces Round #675 (Div. 2), which will be held on Oct/04/2020 19:05 (Moscow time). **This round will be rated for the participants with rating lower than 2100.**

The problems were prepared for you by andrew, hloya_ygrt, AleXman111 and Vladik together with me. We think that we have prepared good problems for [contest:297213]. After that we selected the best ones for this round.

The company Andersen has been holding a competition for the second year already, which is primarily intended to support students of regional universities in Belarus and Ukraine (starting this year).

First of all, we would like to thank MikeMirzayanov and everyone involved in the development of Codeforces and Polygon platforms. No less thanks to KAN for coordination — thanks to him you will be able to understand our problems. And also to all our coaches and parents who taught us to do everything that we can do.

The score distribution promises to be like this: **500 — 750 — 1000 — 1500 — 2000 — 2750**.

Good luck and clean code to everyone!

**UPD**

Congratulations to the winners of the official standings:

1. Yukikaze_

2. lunabbit

3. kamer

4. Potassium_Fan

5. 2018LZY

And overall winners:

1. awoo

2. dlalswp25

3. tfg

4. Heltion

5. hank55663

The editorial will come later.

**UPD**

**The editorial came.**

Notice the unusual start time:)So you want to say what?

Overrated, downvoted

UGHHH, Came to comment the same

Will the two contests be held at the same time?

Yeap, but [contest:297213] will be officially for students of regional universities Belarus and Ukraine only.

This contest starts

1 hour 30 minuteslater than the usual time.Comon, I was typing when the first guy put the message and it went unnoticed to me and i get 22 downvotes, nice.

Looking forward to this..

Thanks, this is a most excellent time for those of us on the American west coast (9 AM)!

Yeah, same here, 10AM for me :)

Its worst for IST 9:35 pm

Wait, where are the testers?

Perhaps they both prepared the problems and tested each other's problems :)

Are problems in proportion to scoring distribution? Or we are going to see another Codeforces Round #657 (Div. 2) where problems were much tough than their scores.

my experience says it will be similar to the one you mentioned.

Query related Score DistributionCan you tell me How score distribution decided by authors?

Absolutely it was not like Round 657, but the scoring distribution was not standard. I am not sure why people strictly prefer downvoting comments, I was just talking of a possibility which came almost true for this round.

Notice,unusual start time. Looking forward to the contest :)

As a tester, I strictly recommend you to partisipate this round. Also I want some contribution.

And Now you are getting contribution for your efforts. Enjoy! :)

Good luck and

clean codethanks :)

"Good luck and

clean codeto everyone!" Is this a hint for Implementationforces.Longstatements orshortstatements ? :pIn China, I have to stay up late to participate in this contest. Hope my rating++.

Previous timing was better 8:05 pm(IST).

October 4 and 5th are two special days for competitive programming contests:- RuCode Championship, BSUIR Championship, Bubble Cup 13 — Finals etc.

I wish to join in but 00:05 UTC+8 is too late for me. Anyway hope everybody has a nice contest.

Good luck and high rating to everyone!!!

00:05(UTC+8) is a bit late for the Chinese contestants.

*Asian

Asia as a continent has a lot of timezones — combine russia, china, Japan, Mongolia and india

Valorant statements again?

Jokes over, you're dead!

could you make my contribution from negative to 0 ?

I know I would add more negative from this comment but still I wanna give a try.

wtf how did this work

oppsss i got earlier here. 1.5 hrs still to go for contest

It will end at 12.04 am here. Hope I can have a sound sleep after completing a good number of problems. And good luck for everyone!

It will be 12.04 am ig

Hey Dear codeforces, first thank you for developing foremost and prime algorithmic platform and I know this process will continue, and second, I hope the ratio of codeforces's contest holding per days will grow to meet one contest every two sequential days. finally, I appreciate your efforts for creating a contest and hold it in codeforces's website:-)

Wishing a good contest without any bad issue and best of luck everyone

Best of luck guys!! Wish you all a good contest :)

Why am I not able to register?

same problem with me

so bad

kingmanas is openly cheating in this contest, I have proofs Manas what's the use of rating when you get it through cheating!!?

Woah! Could you please prove it... Might be a hacking attempt or you are just upto defaming me. If the latter case, then you might have read the article on 'evercookie' too. Mike can easily find you, if he wants to.

Hahahah nice try on scaring me off. No issue, enjoy this fake rating you earn through your cheating. :) I'm gonna expose you soon don't worry

Haha, I know that was meant to distract me mid-contest... Why do you feel the need to do that to me? You sent a DM about some whatsapp screenshots, please forward them here.

The guy who sent me the screenshots doesn't want you to know that he shared this with, so my hands are tied. So you're saved for now. And I really don't need to distract you lol, I'm a CM for the last 3 months so I don't really care But I just don't like when someone like you cheats and gets huge delta

Your statement disproves itself. If a guy helped me cheat on WhatsApp, it means I already know who he is. Whatsapp is not anonymous. Please feel free to share any screenshots you have, haha. One thing your comment does prove is that you are not a CM.

I'm a CM from your college only...hahah

LOL THREAD.

Hey CM, I just came across http://www.cybercrime.gov.in/ and wanted to tell you that sometimes you offend the wrong person and later regret it for life. 12hrs is all you have. Apologize to me from your real account on DM and I will forgive you. If you know me in real life, you also know I don't joke much. How can you be so careless that you messed with me? And if you don't know me in real life then I promise you, you won't forget me until your wedding night.

B should have been 1000 and c 1500

how to solve C?

Consider the contribution of each digit to the sum. For each digit count the number of subarrays before it which were deleted. In this case the contribution is fixed. Consider 1,2,3,4 and the contribution of the 3 for all subarrays before it which could be deleted. The contribution would be 30 * (number of such subarrays) which is 30 * 3.

Then count the contribution for the subarrays after it which can be deleted. This required some weird sum manipulation. Consider 1234. Consider the digit 2. There can be one subarray of size 2 after the digit 2 which can be deleted and 2 subarrays of size 1. So, the contribution of 2 would be 2 + 20 + 20. This can be written as 2 * (1 + 20). If you do this, you'll see a pattern.

can you explain the pattern cause I tried but couldn't find one.

So, for the subarrays deleted before the current digit which we are considering, there is no pattern. The contribution is constant. For the subarrays which are deleted after the current digit, consider a number like 12345. Let's consider the contribution of 2 for every subarray after the digit 2 which is deleted. We can delete 1 subarray of size 3, 2 subarrays of size 2, 3 subarrays of size 1. So, the contribution is (2 + 2 * 20 + 2 * 300) or 2 * (321). Similarly, for the digit 1 the contribution would be 1 * (4321).

Could you please explain it to me a bit in depth please

nevermind

Consider the given string in reverse. Now for each position $$$i$$$ from 1 to $$$n$$$, let us calculate the contribution of $$$s_i$$$. In case the removed substring was to the right of it, then the contribution of $$$s_i$$$ is $$$10^{i-1}s_i$$$. The number of such substrings are $$$n - i + 1 \choose 2$$$. In case removed part is to the right, then the contribution is $$$s_i\sum_{z=1}^{i-1}z10^{z-1}$$$. The summation can be maintained while iterating over the string.

https://medium.com/@harryjobz/bargain-a6cf0b9a5262 I have explained it here since the actual explanation is too long to write it here

Some one please tell me what is the procedure of B I try for 1.5 hour but failed on test case 2

yes please!!

Basically for a particular cell $$$a[i][j]$$$ the palindrome squares which should be equal are

(notice this works even when these are 2 cells and not 4), now our task is to chose an $ x$ such that $$$|x-a|+|x-b|+|x-c|+|x-d|$$$ is minimized where $$$a,b,c,d$$$ are the four cells mentioned above,its easy to see that if $$$a\leq b\leq c\leq d$$$ then any $$$b\leq x \leq c$$$ minimizes this expression; so for convinience pick the average of the middle two numbers.

Here is my submission 94697945

Damn man! I took the average of all four numbers

oh shit i just take all four but check pair wise instead of taking median of four

why average of 4 numbers will not work?

take the four numbers to be: 1, 1, 1, 1000000

got it thanks

But what's the theory behind it, Why such thing occurs? Why average not suitable here for all four numbers??? I'm noob, please explain...

Finding the value where the sum of diffs to some other values is minimum possible is equivalent to following problem: Find a point on a line with sum of distances to some other points on the line be minimum.

Assume some point x on the line, and the number of other point on the line left of it and right of it. If you move x in one of both directions, then the sum of distances changes by the number of points left/right. The sum increases by the number of points on the side you not go to, and it decreases by the number of points in the direction you are going to.

So, to minimize the sum, you go to the direction where more points are...as long as there are more points. At some step there will be same number of points left and right. There is the optimum. And this point is definied to be the median.

Note that with an odd number of points there is exactly one median position. With an even number of points all position in between the two "middle" points are equal optimal.

you can take second largest number as x that worked for me.

It works!if a≤b≤c≤d then Min{|x−a|+|x−b|+|x−c|+|x−d|} => d + c — b — a

Yes, it does work! Could you please explain how you reached this result? All I can comprehend is that you expanded the mod by assuming that the answer lies midway b and c...

|x — a| + |x — b| + |x — c| + |x — d| = x — a + x — b + c — x + d — x = (d + c — a — b)

Why average of b and c can u pls explain why its like taking median of a,b,c,d gives minimum

My Approach was a[i][j],a[m-i+1][j],a[i][n-j+1],a[m-i+1][n-j+1] should be equal

Thanks for your idea bro

I'm not sure if this is the best solution, but here goes

Iterate through the entire array. If for all elements in the array, ar[i][k] = ar[n-i-1][k] = ar[n-i-1][m-k-1] = ar[i][m-k-1], that means that the array is a nice array (this is derived from the definition of a nice array)

So we iterate through all the elements in the array, and we have to make adjustments to the four elements that I listed above in order to make it a nice array.

Call the four elements in question, a_1, a_2, a_3, and a_4. Let the value that we want to change all the four elements to be x. Clearly we want to minimize |a_1 — x| + |a_2 — x| + |a-3 — x| + |a_4 — x|. This is a pretty standard greedy algorithm problem, and the minimum ends up being achieved when x is the

medianof a_1, a_2, a_3, and a_4.My Solution: 94693568

why median pls explain to me. im noob

Please check the reply below :)

Can you please explain why median is giving optimal minimal?

see this!

If array is a[n][m] , for each i,j these 4 numbers have to be equal, namely a[i][j],a[n-1-i][j],a[i][m-1-j] and a[n-1-i][m-1-j]. To make these 4 numbers equal in minimum operations, all the 4 nubers must be converted to the median of the 4 numbers. This is the core idea.

Also, you have to handle cases when n is odd and you are on the middle row or m is odd and you're on the middle column or both.

Thanks everyone i found my mistake

and my code got accepted

What were the hacks on E?

I was printing last two letters wrong in some cases and got pretests passed, so i think that was not covered on pretests

my first submission fails on "bbzzbzzba"

Mine fails too. Should have taken the announcement seriously but this should have been in the pretests.

For the first time I'm wishing someone would have hacked my solution, Could have become Master :(

Some fail on "bccba"

`yzzyx`

From my experience, there were 2 things to consider:

1. You can't remove pairs which are not continuous in the original sequence

2. You need to keep the two stacks in sync, the one with the last different element and the one with the current answer. If you pop from one, you need to pop from the other too. Similarly for pushing

Mine broke on

`aaffaihe`

Also

`bbccbac`

Ashishgup were you able to figure out the hack for your solution?

I am asking since my solution had the same wrong answer on the same line in the same testcase so I was just wondering if you figure it out do reply. Thanks

What is up with MLE on D? M is upto 1e5 that should be doable in 256 mb?

Is there an elegant way to solve C without keeping track of all the edge cases?

think dp

is the contribution of each position i : summation(i) * summation((10 ^ (n — i — 2)) * (n — i — 1)) ?

I'll do you one better, think CP

Is "write a DP" elegant enough?

My DP state is

Transition is to consider two options: to remove the current digit or to keep it. Code: 94671611.

My approach was, take every suffix of the string, now we have to combine it with every prefix(there should be gap of atleast one character between ending of prefix and starting of suffix),maintain sum of prefixes.Combining is multiplying length of suffix with the sum of prefixes and adding it to current suffix.

94711980

Can you explain this in detail?

I think the code is self explanatory, go through it and feel free to ask me anything you didn't understand

My approach didn't have much edge cases except for the first and last digit. You can calculate the contribution of each digit in answer by considering 2 cases :

1. If a substring before the current index is removed, it won't affect the place value of the current digit, say $$$d$$$ and is equal to $$$d*i*(i+1)/2*10^{n-i-1}$$$.(0 based indexing)

2. If the substring removed is to the right of the current index, it's contribution is : sum of $$$(j)*10^{j-1}*d,$$$ where j is from (n-i-1) to 1.

My implementation.

SpoilerI am not able to decode why are we multiplying with i*(i+1)/2. Can you explain a bit??

That's just number of ways to pick a substring in a string of length

ihttps://medium.com/@harryjobz/bargain-a6cf0b9a5262 I have explained it here since the actual explanation is too long to write it here

How to solve D ?

I don't have a proof but... I built a graph:

vertexes are the given points

connect start and all "movement points" with weight: min(abs(s_x — x), abs(x_y — y))

connect each "movement point" with 4 the closest "movement points" by x and y.

And then run Dijkstra on this graph!

Seems if you want to jump from "movement point" A to "movement point" B you have to spend the same time as if you visit all "movement point" between them.

It should be clear why running Dijkstra on complete graph is correct. To prove that running it on this "pruned" graph is sufficient it is enough to show that any "non-adjacent" edge can be represented as a sequence of "adjacent" edges, e.g. if you want to get from vertex A to vertex B and the optimal move is to go over coordinate x, you can as well consider all vertices sorted by x and sequentially go through all vertices that are between A and B in this sorted list.

Hey, could you elaborate on how you proved that you could prune it this way? I was able to prove that djikstra on the complete graph works, but that sol would give a TLE/MLE. What exactly do you mean by ""adjacent" edge and "non-adjacent" edge"?

I call two vertices "adjacent" if they are next to each other in the list sorted by x or by y (so for each vertex you have a total of at most 4 adjacent vertices — 2 by x and 2 by y).

I wrote in my comment above why this works. You may visualize it with 1D version of the problem where the claim will be more intuitive.

Yeah, makes a lot of sense now, thanks!

shortest path problem

I got that there is some prefix and suffix stuff and modulo of long strings involved in C..but how to solve it in linear complexity?

For each number at ith index, try to see how much contribution it can give to final answer.

Observe the

pattern:S = "12345"

Removable Sub-Strings :

length_1 = '1', '2', '3', '4', '5'

length_2 = '12', '23', '34', '45' length_3 = '123', '234', '345'

length_4 = '1234', '2345'

length_5 = '12345'

Strings Left after removing the above Sub-Strings :

'2345' (after removing '1')

'1345' (after removing '2')

'1245' (after removing '3')

'1235' (after removing '4')

'1234' (after removing '5')

'0345' (after removing '12')

'0145' (after removing '23')

'0125' (after removing '34')

'0123' (after removing '45')

'0045' (after removing '123')

'0015' (after removing '234')

'0012' (after removing '345')

'0005' (after removing '1234')

'0001' (after removing '2345')

'0000' (after removing '12345')

'0's are appended at the first for the ease of understanding. You can ignore them.

Contribution of every index in the total

sum:0th index ('1') = 1 * (4*pow(10,3) + 3*pow(10,2) + 2*pow(10,1) + 1*pow(10,0) + 0*pow(10,4))

1st index ('2') = 2 * (------------------- 3*pow(10,2) + 2*pow(10,1) + 1*pow(10,0) + 1*pow(10,3))

2nd index ('3') = 3 * (---------------------------------------2*pow(10,1) + 1*pow(10,0) + 3*pow(10,2))

3rd index ('4') = 4 * (------------------------------------------------------------1*pow(10,0) + 6*pow(10,1))

4th index ('5') = 5 * (------------------------------------------------------------------------------10*pow(10,0))

Hope it helps !!!

Yes got it, Thank you very much for helping me out !!

Am I missing some simpler solution, or was online part of problem F simply

"take offline solution with a segment tree, add persistency on top of that to make it online"?The memory limit is tight for segment tree with persistency. 1024 MB may be better.

Oh, I didn't think about that. I'd say this fact doesn't make the problem any nicer, although others may have a different taste.

I finally got to use the fact that C++64 pointers are larger by 4 bytes compared to C++32 pointers, rip people that can fix MLE by changing compilers.

Can you tell more about which compiler's pointers are 4 bits larger and why it helped in implementing the persistence?

Oops, sorry. I meant 4 bytes. C++17 64 bits has 64 bits pointers I'm pretty sure. It helps in this problem because memory is a bit tight, changing compilers made barely MLE turn into 300MB.

I was able to get a smaller memory footprint in the 32bit compiler by doing two ideas:

[80MB Memory | 1500ms]Use a persistent BIT rather than a persistent Segtree. (https://codeforces.com/contest/1422/submission/94824471) Basically using an array of maps<int, int> rather than an array of ints and use --upper_bound(version) to find the greatest version <= to the one I'm looking for.[100MB Memory | 750ms]As I was making multiple updates for each version of the Segment Tree, I made a small trick to skip updating the nodes in the same version multiple times. The trick was to add a version variable to each node and pass the new desired version as an argument to the update function. That way when I try to create a copy of the node I only do it iff the desired version I pass is different than the version in the current node. (https://codeforces.com/contest/1422/submission/94825863). That way even though I made multiple updates for each version, I avoided creating multiple times the common nodes among those updates. :DI think, it's to cut off solution with algorithm Мо.

I know how a persistent segment tree works, but I'm not too familiar how to use it. Can you explain how you used a persistent segment tree in F?

The problems were really interesting and amazing,but the score distribution is very low for these difficulty level problems

I thought everything other than A was really difficult

I dunno. Maybe slightly more difficult that a typical div2

div 1.3

NOt completely div ,but much harder than a general div2 round

div*div1

another horrible contest

The contest was not horrible. Your participation was horrible

maybe not horrible ... but more difficult than usual

really bad score distribution. E was more easier than C(atleast to me). imo the only reason E had less submissions is just because it was E.

I agree, I spent too much time on D(which I couldn't solve). I missed to submit E by just 1min. By correct scoring distribution we could have seen atleast 600 people solving E.

Lol, I would take my words back "we could have seen atleast 600 people solving E" after seeing number of solutions getting failed in system testing.

I wasn't able to understand E

`In the second example, for the longest suffix "abbcdddeaaffdfouurtytwoo" choose three pairs (11,12), (16,17), (23,24) and we obtain "abbcdddeaadfortytw"`

Why (2,3),(5,6), etc weren't selected? I apologize if I missed something?

Because that is not lexicographically smallest strings

choosing (2, 3) will give string "acddde...." which is lexographically larger than "abbcd....."

Much easier if you don't read the statement, amirite?

gotcha

The difficulty level of this round is much higher than a div2 round

D is similar to Timus 1205

Guys, did you all even face server slow done in last 10 minutes? I tried to go for hacks, but the main site itself wasn't loading. I am doubtful that this round shall be rated, if its the problem for many.

in my computer, all of things is going fine

In C answer is ((9*10^r*r-10^r+1)/81+(10^r*l*(l+1)/2))*s[i] fo i=0 to len(N), where r is element count in right l is element count in left.

In D sort fast positions by X and by Y, a element has edges to x-1,x+1 and y-1,y+1 (closest x to right and left and same for y). Then do a djikstra from starting position. Initial distances are min(s.x-i.x,s.y-i.y). The answer is minimal dij[i]+abs(f.x-x[i])+abs(f.y-y[i]).

A was harder than any of them :(

System testing results on E are such a massacre that it probably would've been too much even for good old days when having CF problems that allow hacks wasn't frowned upon :)

Before system test

After system test

Can problem C be solved using Dynamic Programming?

Yes you can calculate an AGP by dp see the below comment of mine!

One more dp solution for problem C: 94861783

I store 3 things:

value of substring that ends in current pos and has no crosses

(count, sum) of substrings that end in currend pos and last digit is crossed

(count, sum) of substrings that end in current pos and last digit is uncrossed, but it already has some cross before

1 transition is just prev*10 + current

2 transition: we can cross current digit only if we have previous digit crossed, or if we take substring without any crosses and put first cross now, hence count is 1 + all that end in previous digit and last is crossed, and value is just sum of same things: value ended in previous digit and all sum of values that crossed in prev step. Since we don't add anything to the end we don't need to multiply.

3 transition: we can add new digit (uncrossed) to any sequence that ends in previous digit and last is either crossed or not crossed (but already has some crosses before). (it's sum of 2 and 3 in dp) also we multiply it both by 10 and add current digit respective number of times. Because if we have 3 sequences that end in previous digit and we can add new digit to all of them and and this will create 3 different combinations.

And the answer is sum of all sequences where last index is crossed and last index is not crossed but it has some crosses before.

I'm not sure that it helps anyone, but I was so glad this solution worked, so I couldn't help but share with you guys.

C is simple if you make AGP sequence using dp. I got this during last moment but implementation finally completed 5 minutes later Sad lyf.

For ith digit in the sequence how many times it is included is simply.

(i*(i+1))/2 *10^(n-1-i) *d[i] where d is digit sequence + .

(1*10^0 +2*10^1+ 3*10^2.............(n-1-i)*10^(n-i-2 ) )*d[i].

I hope the intented solution is similar.

PS : ACCEPTED ! 94711408

What is agp?

Arithmetic-Geometric progression

Arithmo Geometric progression. For example.

1*10 +2*100 +3 *1000+....

https://medium.com/@harryjobz/bargain-a6cf0b9a5262 you can remove agp from your solution. I have explained it here

Can you please explain how you got this AGP for the right part of every ith digit, I tried to make a formula but that's completely different from this, I thought that if I am at particular index 'i' then contribution of this digit will be

i*(i+1))/2 *10^(n-1-i) *d[i] where d is digit sequence + .

(zC1*10^(z-1) + zC2*10^(z-2) + ... + zCz*10^0)*d[i].

Not: here 'z' refers to s.size() — 1 — i

Can you please help me in getting why my second part of formula is wrong? According to me there should be zCx ways to remove x elements from the given number of elements will it not be correct here?

You can only remove a

contiguoussequence. Hence zCx would not be correct.Ohh yaa....what a stupid thing I was doing...thanx for telling bro!

Similar to what I did in-contest, but I think my approach might be quite a bit cleaner:

My submissionwhy i*(i+1)/2?

Number of substrings till index i is i*(i+1)/2; and if you remove any of these substrings from the front of digit i then its relative position will remain same that is 10^(n-1-i).

For example . for 12345 and I consider digit 4.

if I remove 1 length substrings . which are possible, 1,2,3 If I remove 2 length substrings before index of digit 4 these are 12 , 23. If I remove 3 length substrings these are 123 (only one.)

and for all these removals the relative position of 4 remains same that is we are not removing anything from back thus 4 gets multiplied by 10^(n-1-i) where is is index of 4.

Thanks for the explanation! I had a bit different approach, that got ACed after i found one of the top rankers using the same approach and fixed my code.

nice pretests on E, F

nice sarcasm on pretests

Thank you, we tried our best.

The reason why we need testers — Codeforces Round 675

Actually this one is trivial. The reason why we need testers is Education Round 95. Though it seems they had a couple of testers.

Can anyone give me an example where

`(a[i][j] + a[n-i+1][j] + a[n-i+1][m-j+1] + a[i][m-j+1]) / 4`

will not be the best choice when both N and M are even.1 1 1 9

n=2, m=2 1 2 3 10

(a[i][j] + a[n-i+1][j] + a[n-i+1][m-j+1] + a[i][m-j+1]) / 4 = 4, which gives a total modification cost of 12, however the solution is actually to change all values to 2, or 3, giving a total modification cost of 10.

Thank you! I got it..

I have done the same at first thinking

AVERAGEis the best and all the 4 numbers should either increase or decrease to the average. But consider this basic case where it will not work if one of the numbers is comparitively large say the numbers are`4, 4, 4, 100.`

The average is 28. So, the answer according to you is

`24 + 24 + 24 + 72 = 144.`

But if you consider

MEDIANwhich is 4 here. The answer is`0 + 0 + 0 + 96 = 96.`

This mistake costed me 45 minutes and 2 wrong submissions.

And this mistake cost me "Not Accepted"

.

Horrible pretests on problem A, I've already seen 5 submissions that shouldn't have passed systests get Accepted. For problems like this, where there are many, many solutions, why aren't the pretests adequately comprehensive to handle this?

Constraints on the number of pretests?

I spent half an hour debugging before I realized that for B, you choose the median of the 4 numbers, not the mean... and I'm in AP Stats. rip.

Yeah, its a while ago I was solving a problem that asked us to equalize all elements of an array where you could only use an increment. I misinterpreted it as equalize using either an increment or decrement. Broke my head for 15 min to either prove that it was equalized to median or most frequent element or avg, and gave up, only to realize I had read the q wrong. But luckily after the contest when searched up upon the misinterpreted problem, I found this — proof

Happy learning!

Can you plz give me an example of elements, where taking median gives us the smaller answer than when taking mean. (I mean choosing the element among 4 elements in problem B)

UPD: Found the example from someone

1, 1, 1, 100

Problem E: Test case 46

`wrong answer 141st lines differ - expected: '430 hhccc...io', found: '428 ccccc...io'`

Isn't my string lexographically small ? 94701491aropan can you please explain?

Got it. Thanks

This suffix starts with

`hhhhhhhxxxxllllllhhhhhhhvvllllccccccodd...`

. You cannot delete all`h`

(because odd group) before block with`c`

.when will official tutorial come?

In the 2nd test case of 2nd question if we convert the matrix to 4 3 3 4 5 6 6 5 4 3 3 4 its still following palindromic condition and we require 36 operation and 36<42 then is there some error..plz help

The last element is $$$18$$$, but not $$$12$$$.Even if you use your matrix, the answer is still 42, not 36.

Wrong answer on problem E:

`wrong answer 141st lines differ - expected: '430 hhccc...io', found: '428 ccccc...io'`

Excuse me?

Exactly.. How?

bccba

Correct answer: 3 bba ...

you: 1 a ...

nice catch

Thanks!

Can you please help me with possible hack for test case 42? From the comments I can see possible hacks for test case 46 are correct from my solution.

Check for "bxxb", you will understand that we had to do the operation only once. The answer for this is -

2 bb

1 b

2 xb

1 b

I am afraid of the round which affiliated with some local CP competitions.....

One of the worst div. 2 contests i ever written. Very huge gap between problems C and D. And problem F, seriously? Basic ST problem only for write two different STs without any idea, it is problem for educational round not for competitive. So upset that I only read this problem 7-8 minutes before the end and done it 30 seconds later contest ends :'(

P. S.But the lesson is learned: to read all problems, even F, even in div. 2, even when you didn't solve D and E

Can anyone please tell me what is wrong in this code for div2-C problem? i got WA on test 6.

94699300

Instead of this

`ll c1=((ss[i])*x)%mod; c1=c1*i*(i+1)/2;`

Try following

`ll c1=((ss[i])*x)%mod; c1=c1*((i*(i+1)/2)%mod);`

Thanks a lot! It worked!

I wonder why the above comment was down voted.

`wrong answer 141st lines differ - expected: '430 hhccc...io', found: '428 ccccc...io'`

can someone explain? Got it.explained in above comments.Answered above https://codeforces.com/blog/entry/83360?#comment-706838

Got a wrong answer on test 46 for problem E.

`wrong answer 141st lines differ - expected: '430 hhccc...io', found: '428 ccccc...io'`

How is the expected answer correct here? (`hh`

should have been removed, right?)Just what I came here to write

Perhaps the string looks something like

`hzzhccio`

. In this case, removing the z's puts the two h's together, but there is no way to remove the h's.Can you please help me with possible hack for test case 42(not sure why my solution failed at 42, 4711203 )? I can see above potential hack for test case 46 seems to be correct in my solution.

Output for "hzzhccio" :

6 hhccio

5 hccio

6 zhccio

5 hccio

4 ccio

3 cio

2 io

1 o

If there were letters in between hh that got removed, then hh cannot be removed as you can only remove characters are that

originallyadjacent.Can anyone tell me why there are so many 'Wrong answer on test 46' for Problem E (including me,T_T)?

Many people over looked the fact that you can only remove letters that are ORIGINALLY adjacent from the original suffix meaning that for something like abba, you can remove bb to get aa, but since aa was not an originally adjacent pair you cannot remove it. It is kind of surprising because I think it was very well explained in the problem statement.

I have paid attention to this, my answer does not produce the error for type 'abba'，but I still get a 'Wrong answer on test 46'

I just looked at your submission for E and it seems that you printed out an incorrect length

I get my answer dont update the 'last' char in some situation

If you get WA on test 46 of problem E, this case may be helpful:

Spoilerinput: abba

output:

my ansser is right for this case, but still get a 'Wrong answer on test 46'

My code solves that case (except that the 0 should be a 2), but I still get WA on test 46.

F for the bois, who thought mean would give the minimum number of operations instead of the median.

Problem D

in the input line, should it be :

instead ?

no

yeah realized now, its correct, sorry for the noise

I did a quick explanation of problem D — I don't do many of these, so let me know what you think!

https://www.youtube.com/watch?v=oJcU9RWnoU0&feature=youtu.be

(tldr: look only at "adjacent" nodes instead of full graph)

94699532 why I am getting wrong answer on pretest 2 in Problem B ?

Anyone explain me

If you have a collections of numbers, and you want to make all of them equal. The minimum operations would be achieved if we make each of them equal to their median, not the mean.

please explain. why median not the mean ?

There is a mathematical proof for that. You can search the internet for the formal proof. It is apparently very well know technique. Also, if you don't want to delve into that, you can ternary search because the function of minimum operations form a V shaped curved where the median is the local minimum. Actually take this case for some self convincing, Credits to someone above. 1, 1, 1, 1000000

OK, I got it

Thanks brother

You are taking average of the 4 values, so it will be round down integer value, but it is also possible that optimal answer exist when you take it round up.

so should we take the median of the middle two numbers.? how should we do that

In Problem D , can someone prove why just connecting instant location to all four adjacent sides will work ?

(Incorrect sample, this only works for point->lines or point->rows or line->line or row->row... not point->point) Image that you are on the spot (0,0) and have 2 instant-moviments at spots (3,5) and (7, 8). If you add all forwards edges: - (0,0) -> (3,5) = 3. - (0,0) -> (7,8) = 7. - (3,5) -> (7,8) = 4.

You can see that for all triple path xyz (with x->z greater than x->y), the total length will be x-> y + y-> z.

Then, you only need add edges to the nearst spot.

(3, 5) -> (7, 8) = 3 not 4 you only need to move 3 steps in the Y coordinate.

Yes, you are right.

My sample is incorret. But this ideia is corret when you use:

point(0,0)->row(10) = point(0,0)->row(5) + row(5)->row(10)

Can anyone prove why the answer is the median of the 4(or less) elements in problem B?I googled it and used the function from here, but it doesn't explain this very well.

You could have also tried ternary search, the function looks like a V, where median is the the local minimum.

This will help — my other comment . The second proof in the stack overflow page seems more intuitive.

Try proof by contradiction. Try making every element equal to a number to the right of the median. You'll see that the cost is more.

Just try doing it for 1 100 101 102, It should be obvious

Imagine that you have a ordered array V of N no-negative elements and will transform all elements into Y=0.

ohhhhh, thanks, i understand now

imagine you have 4 points sorted and listed on the x axis p1, p2, p3, p4 It can be proven that the best point let's call it p0 to make the distance from all 4 points to p0 minimum is either p2 or p3

First it's obvious that p0 can never lie somewhere less than p1 or greater that p4 Second let's assume that p0 is somewhere between p3 and p4, let x = p0 — p3 (the distance from p0 to p3) the total distance is: (p3 — p1) + (p3 — p2) + 3 * x + (p4 — p0) = (p3 — p1) + (p3 — p2) + 2 * x + (p4 — p3) now imagine p0 is at p3 the total distance now is: (p3 — p1) + (p3 — p2) + (p4 — p3) we reduce the total distance by 2 * x similar proof can be done if p0 is located between p1 and p2 try it on paper and it will make more sense.

Just loop over the four numbers and take the one which we can get with minimum moves.

I know that, I claim that the two numbers in the mid after sorting are the best answer :D

Thanks for these quality problems :)

Problems were very educational, Thanks.

Me: I have been doing competitive programming for 8 years

Also me: Use average instead of median in B (fortunately it was considered on the pretests c:)

Weak pretests might leads to huge destruction in B.

Can someone tell me why this submission for F gives runtime error? I can't seem to find any overflows or index out of bounds

Maybe your LCM function will overflow with large a and b. I think you have to cast to long long in that function

I use

`#define int long long`

, I thought it should cover that? Anyway, thanks for looking into it:)Yes, I missed that. It seems that in your ST query function, you get the answers of the two queries in tmp. These could be large. Then in the lcmModuloM function, they could get out of bound of the array primes. Also the LCM function doesn't work nicely with modulo. So

`LCM(LCM(a,b)%M,LCM(c,d)%M)%M != LCM(LCM(a,b),LCM(c,d))%M`

Oh I see, thanks. So how can I calculate the LCMs properly under modulo operation? I'd read another comment saying 2 segment trees are required...

aropan Can you please help in providing test case 42/46 for problem E? If I understand correctly many people are struggling in figuring out that. Even from the comment section I am unable to find any suitable hacks for my solution. Would be very helpful if both the test case can be broken down in some smaller string and shared in the announcement or editorial.

check this case baaddabdaa. your output 4 babd 3 abd 4 aabd 3 abd 4 dabd 3 abd 2 bd 1 d 0 1 a correct output 6 baaabd 5 aaabd 4 aabd 3 abd 4 dabd 3 abd 2 bd 1 d 0 1 a

Thanks, it is useful.

my code is giving correct output on this test case but it is still failing on test case 46

Screencast of somehow managing to submit n^2 for C and failing E, with solutions for at least A-D

whatever, this comment will be buried anyway

Can someone plz explain where my code for

problem Cwent wrong, i was able to pass the sample tests given in the problem. Thanks in advance and i wish you a +120 rating delta the next contest!SpoilerHave you considered the possibility of overflow in the expression containing i*(i+1)/2?

Now i can die in peace , thanks man!

Bruh... My soluon for F got TLE during the contest's systests and AC during the upsolving. I didn't change a thing! Look at my two last submissions.

Your solution have 2979 ms time. TL is 3000 ms. I think 1% error in time is not critical for testing system.

After debugging problem E for around 2 hours it's finally accepted :|

Yay me -_-

I'm totally surprised that my code for problem E passed pretests, and also worked for the first 85 tests in systest. I had at least 10 errors and typos (HUGE errors). I still can't believe what happened. This is what I get for using hashing, binary search and sparse table in a problem that can be solved using a stack.

What complexity was C supposed to be? I found an n^2 (n being number of digits) solution but didn't bother coding it cause I figured the complexity was too high.

n or nlogn would work.

i think you mean n here to be the number of digits which will happen to be around 10^5 cause n is 10^10^5 and that can lead to i don't know how much time to compute . If we process the digits from right to left and take the contribution of each of them . It would be O(number of digits ~ 10^5) then :)

Problems were really good especially C. It was that type of problem which i had not solved in the past. Thanks codeforces for such contest. Hopefully , I am becoming specialist according to codeforces rating predictor.

Thanks for the round! Finally got Div 1 :D (exactly on edge, 1900)

I am also exactly on the edge(1899) :(

In problem A, why isn't a+b+c-1 a possible answer ?

Integer Overflow:P

It is! I've seen many ACs with it.

Yes it is. Integer overflow got me here. :(

Although my solution for problem B is O(n*m), I'm getting TLE on test case 6. Please could someone go through my code and identify the bug. Thanks a lot in advance!

Code

may be the traversal of vector is the reason.

But I haven't used a vector. Thanks for your code anyways

vector/array

Idk why, I submitted the same code now and it gave AC .

https://codeforces.com/contest/1422/submission/94691338 here's my submission with same approach.

94730877 Problem B Can anyone help me to find the bug in my submission, thanks a lot~

Between the 4 values, find the median, not the average. Finding the median gives you the minimal answer. For example, 1 1 1 10. If you use the median, 1, you only need 9 operations. If you use mean, 3, you need 13 operations. I made a similar mistake lol, hope this helps.

oh, i got it. thank you, bro

Can anyone explain the problem A? I don't understand what it mean by "no three of its corners lie on the same line, and it does not cross itself".

In the first case, if 3 points are collinear, then you cannot form a quadrilateral by adding 1 extra point. I think the second is to prevent concave quadrilaterals.

If you search in Google "queries in range of lcm modulo", you can find a Codeforces blog in which an user asks for almost the exact statement of problem F from this round, and provides a link to the original problem

Actually offline version can be solved with Mo's Algorithm also.

can we solve problem A by trying to give it a shape of trapezium? ceil(sqrt(a[0]^2+(a[2]-a[1])^2) a[] is sorted.

This should work. The initial distance is sqrt(a[0]^2 + (a[2] — a[1])^2). Now, pivot the longer side away from the shorter side. The final distance will be sqrt(a[1]^2 + (a[0] + a[2])^2) when the longest side is parallel to the the shortest side. The final distance is greater than the initial distance. So, we have a continuous function which goes from the initial value to the final value. So, any integer value between the initial value and the final value should work. We may also need to show that the difference between the initial and final lengths is at least 1.

https://codeforces.com/contest/1422/submission/94669432

https://codeforces.com/contest/1422/submission/94746685

A slight modification to your submission. Hope it helps!!

Thanks man! This cost me 2 wrong submissions.

When will editorial be posted?

Unofficial one for A to D problems. AtoD

Can we solve problem C something like this?

I know these are incorrect transitions , but can we solve this problem in this way (using these states)?

I just did this. dp[n][3] means the sum of the current state, also we need to maintain another array cnt[n][3] means how many different way are currently at this state, so we can make the transition by two array. Submission

Thankyou so much for the reply and for the logic as well.

who can give me the solution？thanks a lot

Would anyone check my code for problem D? it's wrong on the 4th test case. 94750634 thanks.

I think the usage of

`mark[x] = 1;`

is wrong. Instead the push onto the queue should be done whenever the relaxation is done.oh, thanks.