Second edition of Round 2 will start at 08:00 on June 13, 2015 MSK.

Don't miss your chance to win a tshirt and advance to the final round.

The round is made by GlebsHP, he even promises to share the problem analysis right after the end of the round. What a nice reason to wake up early on this Saturday!

UPD: The round is right around the corner!

After this contest if someone solved 0 is in 512 will he/she take a t-shirt?!

Are you serious man :P?

People are talking that t-shirts only for indians,please tell us that we can get t-shirts from any place ^_^

Only Indians are eligible to get AC in Yandex, so how would you want to win Tshirt with 0 problems solved?

I always thought that Yandex sounds a bit indian :]

Firstly it said "The contest is over. Submissions are not allowed" and then it allowed to

virtuallyparticipate. Was that supposed to be like that xd (I started)? Please don't do it to me for the second time ; dPlease, press the "start virtual contest" button.

There is no such button anymore.

No such button now :(

There are no standings. (Although that may be because nobody has submitted anything.)

i clicked on it, and it shows "The contest is over. Submissions are not allowed"

Seriously? Surprize virtual contest?

For how long virtual contest will be open?

we'll fix the starting time for everyone on 8:00 in a few minutes

So everyone who started virtual contest earlier will have an advantage?

Was it possible to start it before 8:00? Or do you mean they didn't set start time to 8:00 for people who started virtual contest?

Screw you guys, I'm going home

Is this not a "start virtual contest" button that I ask everyone to press on the title page?

This has no sense: I've pressed 'start vritual contest' button at 8:20 and saw "time left: 1:20" (as if I pressed 'start' at 8:00). This totally ruined contest for me.

Don't get me wrong: problems were great and I'm in top-512 for sure — nothing more to wish as a participant.

But as a human, it was

reallyannoying to see that Round 2.2 went on par with Round 2. In both rounds, there was a non-straightforward way to see problems (random F5 pressing in Round 2, "start virtual participation" in Round 2.2), which put participants in different conditions. And in both rounds, administration neglect that fact and try to continue contest as if nothing happened.In round 2, the announcement about failure was after about 40 minutes, but it was obvious from the start that something went wrong.

This time, there was a fast announcement that one should press "start virtual participation". But it was totally confusing: I came here for a normal round, not a virtual one. Why should I press that button? Is it safe? Will my participation count in overall rating? Will I save my time or it will start from 8:00? All those questions arouse in my head and I was not sure about pressing that button. And immediately my fears were proven: timer started from 8:00, consuming some part of my time. And after opening standings at the end of contest (my time was out), I saw people with 1:33 and 1:28 virtual participation time (meaning they were still competing). Ridiculous.

Yandex.GEOMETRY.comI have no idea about geometry.. i could do only 1 question :(

I have ideas,but can't implement coz have not knowledge how to sort vectors by angle,how to find angle and others...

Here I think,for each bitset of length n(which means use i-th vector or not)we should add all angles in 2 consecutive vectors and it should be 2p and then anwer will be "YES";

accurate!

Am I stupid or D really needed knowledge that every Pythagorean triple can be described as (

d(i^{2}-j^{2}), 2dij,d(i^{2}+j^{2}))?? It took me much time to solve this problem and I felt sad seeing so many accepted submission xd.And how you can solve the problem using that fact?

Just generate all Pythagorean triples and check answer by obvious dp xd.

You can generate all possible Pythagorean triples and then do simple dp.

Can you show your code?

http://pastebin.com/x9ZFM6Gs

Just make graph for numbers less then 10

^{6}: from (i^{2}+j^{2})*k to (i^{2}—j^{2})*k and (2ijk). After that dynamic.-_-

-_-

I think anyone who has done anything on projecteuler has used this theorem about 37 times.

What was in 50th test of problem A?

I apologise for not being able to publish analysis immediately. Hope to finish it before today's GCJ.

I'm very interested to see tests 17, 21 and 31 for problem F. There's something strange about my verdicts.

21 and 31 are too large to be posted.

Test 17: baababbbbaa

Answer for test 17: aabbabbbbaa

What about 4 test in D?

Thank you, found my mistake already. It's so stupid :(

I used the following logic to try A:

Firstly, if 2 vectors are not in opposite directions, then i assumed that the sector formed by the smaller angle between these 2 vectors is reachable.

Hence, using above inference, i sorted all the given vectors according to the angle they make with x-axis and checked between which 2 vectors does our destination lies. If the difference between the angles of these two vectors is less than PI, output YES , else output NO.

This gave WA on test 39. What might be the fault in this solution?

You may have problems with precision. Because coordinates can be up to 10

^{9}, the difference between two angles can be about 10^{ - 18}, so you need to use exact arithmetic.is there any way i might do this with double? i don't have much experience in handling precision errors.. what is the suitable value of eps in such cases?

I was calculating angle using atan2; code with eps=1e-12 gets AC.

So either tests aren't good enough or difference between angles can't be so small.

Tests like this should have very small difference in angle:

For this test, the answer is

`YES`

. And for the next test, the answer is`NO`

:Better way check precision problems is to use vectors like (1000000000, 999999999) and (999999999, 999999998). Atan2 returns the same double for them

Hah, funny. At first thought there shouldn't be difference, because your angle is approximately just 2 times smaller, but what is important here is that atan2 values for eatmore's points are very small, so they are kept with higher precision and that is big difference. Nice!

There were about 10 tests with very small angle differences (about 10

^{ - 18}, as eatmore mentioned). No idea why your code has passed, I will investigate this case in the free time and post the result here.Is there some place I can check my T-shirt address :) to make sure it's correct?

The registration form is inaccessible now.

are you from india ? :D

No. I hope this doesn't disqualify me! :)

I think so :) what's your position?

how to submit my address ? I used social login while registering and it didn't ask for any address then as far as i remember.

I am not sure https://passport.yandex.ru/profile/address

Anyway it would be great to know official version about it.

Did registration form ask address?! I couldn't remember.

27th place after 3 rounds, what a sad story :((

Where do you check place?)

https://contest.yandex.ru/algorithm2015/results/

Thanks!

I'm eager to know how to solve E. If only rectangles nx0 or 0xm would be allowed! Then I will be able to solve it using just one flow. But in order to exclude them I need to run

nmflows and that clearly TLEs. However I got an idea how I can speed it up, but it seems too complicated...I used m*n dinic flow and it was fast enough. The execution time was always below 1 second.

nmKuhns should be fine.nm Kuhns without any optimisation attempts whatsoever runs in 1.037s. I was surprised too!

Could the organizers, please, provide a way to check which address we used when we registered for the contest? Thanks!

Any news about t-shirts?!

Check your email or go to algorithm.contest.yandex.com for more information

Who are the t-shirt winners of Warm-Up round?!