Hi all.

On Sunday, September 7, at 19:30 MSK regular, 265-th, Codeforces round will take place. Problems are prepared by me, Mikhail Tikhomirov. Round will be for both divisions.

Standard (not dynamic) scoring will be used for this round.

**Div. 1: 500-1500-1500-2000-2500**

**Div. 2: 500-1000-1500-2500-2500**

I would like to thank Gerald Agapov (Gerald) for his help in problems preparation, Filipp Rukhovich (DPR-pavlin) and Alexander Mashrabov (map) for round testing, Maria Belova (Delinur) for English statements and Mikhail Mirzayanov (MikeMirzayanov) for creation and development of Codeforces project.

This is going to be my third round on Codeforces, and I tried to make problems as interesting and diverse as possible. Hope you will enjoy this round. Best of luck! =)

**UPD:** round is over. Thanks for participating, hope you liked the problems.

Grats to all the winners:

**Div. 1**:

Special respect goes to simonlindholm, the only participant to solve the hardest problem E!

**Div. 2**:

Editorial is here.

Your next contest after Three years....!

I wish to see more Math.:)

I think you'll get what you want.

:)

I hope for no math.

Why both hope for math and not math problems got negative votes?

Because it's spammed on each contest announcement?

Then explain positive votes in this comment

Edit:it's no longer up-voted (but it was +13)Oh, I hadn't downvoted that one. Fixed.

I meant if it's really spam why people in codeforces up voted that comment ?

Edit:it's no longer up-voted (but it was +13)Dunno, I provided a reason for why downvote (there are dozens of short comments like "I hope there'll X" and "GL&HF" and "thanks for X", and I stopped being interested in reading them after the first 100), not why upvote :D

CF voting is really strange, I have no idea why it behaves the way it does. Sometimes I'm thinking the answer could be 42.

@Xellos I will keep that in mind from next contest.:)

Hey man, I just keep repeating that because this site is not named "mathforces", and others keep making "Hope for math!" comments. (!!)

Oh yeah, I forgot that the optimal method to reduce the "Hope for math" spam is to add "Hope for no math" spam.

While speaking about math, have you noticed that the absolute value of your contribution is a perfect cube? Also the absolute value of mine is a 4th power added to 1 :o. It is a very interesting topic :o.

Here , I have made the absolute value of your contribution a 4th power added to 3 now :D

Actually, it is the absolute value of a 4th power added to 3 :D.

Also your contribution is a fourth power :o.

why? I think Math is difficult

Why? I think math is easier than other!

I love Math.:D

Math is beatiful guys and haters shut up xd. If it weren't for math, there won't be competitive programming xd.

I hope I can do better！

I m going 2 rise

The green knight rises (poke batman).

I think this announcement sets the record for the shortest announcement for a contest ever. Keep It Short and Simple! :)

Keep it simple, stupid! (the KISS principle!)

I like this simple style,Keep it short so that people don't miss part of important message.

Good way to get more upvotes :D

Wish to learn new concepts ,new tricks and new things :)

...

Why is everyone in a hurry?! Contest's blog 5 days before contest?! Once they posted it a few hours before the contest :D Oh, and no thanks to Gerald or anybody else?! What an unusual contest blog...

But higher topic rating than last contest :D

Perhaps this round have something unusual :)

It is my first time to compete, I hope i can solve at least once.

Welcome. I hope you can solve all.

Good luck and have fun! scores up, up and up!

How to register for the contest ? I can't find any link for registration

You will be able to register a few hours before the contest.

Long Waiting For this contest

Good luck everyone

Best of luck everyone...! :)

cheer up!

First time above 1700 and I have a question regarding divisions. My friends take part in Div2 but I can't Register to Div2 due to 1700+ rating.

Will I be able to join as virtual something when Div2 starts without ranking or I must register in Div1?

You can enter virtual Div2 only after the contest is finished

now you can only enter div1

that's bad but thanks.

What type of scoring will be use ?

GoodLucktoeveryonewhy do u combine with normal text?

I don't know ? 'Cause are different topics ? My question and my wishes ?

how to solve

Div-1 B?Bruteforce, I think. There are 6 permutations for each of 7 point (you can fix the first), so it is 6^7 ~ 300k — fast enough.

Or generate all variants for 3 points, build all cube and check points.

How check? Length test&?

How check? Length test?

And how do you check if it's a cube?

I had an idea to calculate all pairwise distances for some fixed permutation (there are 28 segments in total) and sort them. Now, if first 12 of them are equal, next 12 are equal and last 4 are equal, it's a cube. But I didn't have time to code this solution and can't prove it

i'm convinced this idea is correct, but i dont know why my solution 7712723 gives

TLE!Because that's incorrect idea and obviously your solution is wrong. Also, your code-style is very dirty. Learn Math and don't ask stupid questions anymore.

Lol, that's pretty rude. I'd ask you to learn some good manners first.

Oh, I so ashamed...

Один стою в белом пальто красивый.

Map is not so fast thing:) Try to rewrite it without map, it should pass — my solution is basically same, but with vector&sort (7708825), and it works in 0.7 seconds.

my solution 7712723 with the same idea

TLEd! :(Could you please explain how we can fix one point if all of them probably were shuffled?

It would be great if you could write problem A shorter! At the first, I was confused.

Thanks!

How to Solve Div -2 C ?

they key observation is that if we don't have any palindromes of

length 2 and 3, then we will have no palindromes at all (other thansingle characters, ofcourse).How did you arrive at this conclusion? Have you solved a similar problem before or was it intuition or did you think of this during the contest?

intuition. any palindrome of length

n> 2 has a sub-palindrome of lengthn- 2 as its substring (u can get this by removing the first and last characters).easy to note that we can build palindrome by add same character to begin and end of another palindrome. just consider the case of odd and even length, so we could just check every 2 and 3 consecutive letters

I made this observation during the contest but do you still have to loop over all possible strings after the input?

Can u explain ur solution !

Suppose you have a palindrome of length at least 4. Delete the start and end characters. Now you still have a palindrome. Thus you only need to consider palindromes of length 2 or 3. Find the rightmost character that you can increment to something satisfying s[i]!=s[i-2] && s[i]!=s[i-1] (within the bound of p of course), then fill the remaining letters with the "smallest" letter possible.

Damn, I gave up thinking on that because it seemed complicated to check if replacing a character would cause any of the previous strings to become a palindrome.

I got my first hack on codeforces! I tried a little different strategy this round. I solved B and A quickly and then tried to solve C within the first hour. But I couldn't. Nevertheless, I spent the whole of the second hour trying to hack but was not very lucky until the last 10 minutes. About 7 minutes before the end of the contest I managed to hack a solution!

Nice problems!

How to solve D? I was struggling for quite a while, but didn't come up with anything faster than 100 FFTs (or 2 FFTs, but for vector with size 10

^{7}) :(My solution relied on the fact that we only need 1e-9 precision. Basically, with probability 1-eps we never get to very high levels.

I managed to find a suitable algorithm for 464C - Замены в числе, yet couldn't solve it because I couldn't read the input properly. I don't know yet what was causing the quite straightforward input code to fail... Here's my last failing submission : 7713194 If I add assert (c >= 0 && c <= 9); after line 43, the assertion fails...

This was a great contest :) Thank you for writing! I liked many of the Div 2 problems.

I have a feeling that a lot of A-Div1's will fail.

I see some hacks on div1-C, for example Merlininice has quite a few. How can a solution to div1-C be incorrect? It looks really straightforward.

Are you now scared that yours too is incorrect?! :D

Some people did not take the length by module, some took it by module 10

^{9}+ 7 instead of 10^{9}+ 6.But why does one need the length at all?..

I guess you're saying that people were computing "length of expansion of digit X" instead of "10 to the power of length of expansion of digit X"?

Exactly.

My solution got hacked also. It was totally a straightforward dp problem. I just couldn't figure out why it was giving incorrect ans. I thought keeping the length and exponent value would do the same thing but totally forgot that if I kept the length, I would have to do modulo 10^9+6 on the length according to FLT. If only I'd kept the exponent value instead :(

10^9+7 instead of 10^9+6 ????

problem statement say 10^9+7

If we store just length, we're computing 10^len at some point. Using FLT, we can just compute 10^(len mod (10^9+6)) instead, so we can just take len mod 10^9+6 instead.

One should use 10

^{9}+ 6 in order to keep length. Because it is the period of pows (check little Fermat's theorem)New information for me, thanks :)

do you mean that if I saved the length instead of power of 10 and make it modulo 10^9+6 I will get AC?

(applies only if and are coprime)

Hey Petr, didn't you make a similar mistake (taking different value for MOD) in a past contest? I remember reading about it on your blog!

If I remember correctly he made a missprint (10^9+9 instead of 10^9+7 or sth similar), not a conceptual mistake like the one mentioned above.

so do ztxz16 and Kostroma.

C , D div2 anyone ?

I had this algorithm (which I did not implement) for solving 465C - Нет палиндромам!. Is this a good approach?

Treat string as a number in base p.

The function check(x) tells you if x contains no digit > p and that it contains no palindromic substring of length > 1 (both of which can be done in linear time).

Your function check is not monotonic, so binary search won't help you here.

How did you solve it?

Not the contest that I expected and hoped... BTW, Palindromes are AWESOME! Why should somebody be bad with them?!

Here's a Brainfuck solution to problem A (reads and outputs as ASCII). The cells are divided into groups of two. The first cell of the group is 1 if the index is in range 0 to n and 0 otherwise. The second one contains input for that index — either 1 or 0.

The code:

Weren't E div2 — C div 1 pretests very very weak?

I see naive implementation solutions pass pretests!

Wow, really???

Yes, See those for example 7713065 7710477 7710991

I tried to hack this solution

by this case

2 0 0

but it gives me Unsuccessful hacking attempt

with this case variable pb will be used without initialization

how this solution pass my case ?

pb will have junk value, but that would not contribute to answer, just that "break" will not be called.

How to solve div-2 D ?? I first tried to take all 6 combinations from 24 , that is 24C6 , clearly it isn't suitable !!!

When A,B are too easy and others are hard (div2), this will happen: I can see about 1500 rank difference between 2 users while both of them have solved 2 problems. The rank of the participants which solved 2 problems, starts from 400 and ends after 2000. This is a huge difference! I think this shouldn't happen.

Thanks :)

In this contest, if you can solve only A, B which is common with Specialists and low Experts, your score will depend on your time, which isn't kinda bad. If you succeeded in solving C, or even D, E, those tasks will tremendously increase your rating because they are simply made for coders who has ideas and knowledge.

I misunderstood Div1-C problem statement :-( But problems are very interesting. Thank you for preparing this round :-)

For Div2, a and b was easy and c d e was hard. I have solved A and B but I got rank 1100... To many contestants solved just 2 problems and very big difference in same, two problem solvers.

Who else forgot to check if the cube side length is greater than 0? High five!! :'(

High five! :)

I got wrong for DIV 2 D , because I used float instead of integer while printing the cordinates , my programs prints the cordinates in exponent form in one of the test case and is judged wrong.

Did someone else also failed system test in this way before ? Link : 7710844

Why ratings update is delayed too much in codeforces after system test is finished? (while it's fast in topcoder)

the relation between system tests and rating update is inverse. i don't know why .

My guesses:

System tests are delayed — because they need to add the newly added hacks to the final tests.

Rating updates are delayed — because they are figuring out the cheaters in the contest.

Good question, thanks ^_^ waiting for editorials...

"Editorial will be up in an hour or two." is a Infinite Loop statement.

Can someone give some hints for Div1 C?

div2 probC

runtime error. changed to

accepted. life sucks :(

~~1).~~`new char[n]`

is not supported by standart, it's only gcc feature.2). You need

`n+1`

chars because of extra`'\0'`

character at the end.1) LOLWUT?

1). Sorry, confused it with

`char data[n];`

if you can use

`cin`

, why don't you use`string`

?Looks like tests in Div1-B are weak — I mean overflow. I have just written an int64 solution, which use values like maxCoord^4 (euclid length of vector product). And it's full solution. Is it difficult to create test to crash this solution?

I think technically you have calculated a hash of the vector's length (have calced it modulo 2

^{64}) and then you are only interested if two hashed values equal or not, so it seems OK.These submissions are very much alike, aren't they? This 7704944 and that 7706248 And this 7713468 and that 7713301 too!

I'm a policeman:)Oh my rating :))

very happy to get 284 point

can somebody explain div-1 C it is not so clear from editorial

Didn't read editorial, but may be can help

First of all, we can notice that after all transformation behind every digit will stay some sequence. Suppose we know these sequences for every digit [0..9], and for digit d sequence has length len[d] and value val[d]. Now, to compute result for input s, go though s and result = result * (10 ^ len[s[i] — '0']) + val[s[i] — '0'].

Now we need to compute len[] and val[]. Let's use DP technique. Consider len[] and val[] calculated for some suffix of transformations queries [k..m-1]. How compute them for suffix [k-1..m-1]? Use similar ideas were used to compute result. Take query d->t, go through t and value = value * (10 ^ len[t[i] — '0']) + val[t[i] — '0']; length += len[t[i] — '0'];

val[d] = value; len[d] = length;

All these formulas come from Horner's method.

Country wise standings are updated. Sorry for the delay.

Round Stats