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В прошлый раз гарантировалось, что "задача A будет в нем именно та, которую мы считаем наиболее простой". В этот раз это не гарантировано?

:)

133

Thanks so much for arranging the problems in expected order of difficulty.

Вновь жюри порадовали системой оценки и порядком расположения задач, спасибо.

Впервые попробовал взломать чужое решение. Сначала не влез тест в 250КБ. Тогда я сделал генератор – "таймаут 15сек". Пришлось сгенерировать не слишком сложный тест на 218кб – "Невозможно обработать взлом, попробуйте еще раз" – ну что за фигня? (

Задачи классные, хоть я и слил контест... :) Спасибо

Задачи сегодня какие-то очень тривиальные.

Были ли у кого-то кроме меня проблемы с загрузкой страниц во время контеста?

This contest started horribly for me (it took me a very long time to correctly deduce the formula for A, noob :P), but ended as my best placement ever (35th). I'm really happy, and I really enjoyed this entire competition, all the tasks were very interesting! Especially the Web one (even though the idea behind it is not that difficult, the formulation was very nice :D).

Gratz everyone!

First time used Yandex's notepads to solve problem A :D

so quickly rating!

Woohoo. Division 1. да, детка!

hard questions really! specially question A. I hope the editorial will be available soon. And really thanks to all of you who prepared this great contest.

Спасибо за задачи их авторам. Я, конечно, не претендовал на высокие места. Но обидно, что третья упала из-за маленьких размеров массива. И когда я в одном массиве увеличивал элементы и выходил за его пределы, я перескакивал на второй и там увеличивал. А увеличивал я... ответ. В итоге вместо 1 801 1601, я выдал 2 802 1602. Ну хоть 1, 2, 4 не свалились...

in the problem-B forming teams

is it not correct to just find the number of odd length cycles in graph described by 1..n??

sorry for this question(on this blog), but i am intrested why next round time is difference 08/18/2012 11:00AM

UPD: cout << (a+c-1)*(b+c-1)-c*(c-1) << endl;

You can see this submission 2013799.

some tips about problem A

a=2 b=3 c=4 total=18

    O O O O   
   O O O O O
  O O O O O
   O O O O

a=1 b=4 c=5 total=20 X:2 O:18

    O O O O X
   O O O O O
  O O O O O
 X O O O O

a=4 b=1 c=6 total=24 X:6 O:18

X X O O O O   
 X O O O O O
  O O O O O X
   O O O O X X
	   
a=5 b=6 c=1 total=30 X:12 O:18

       X
      X X
     X X X
    O O O O   
   O O O O O
  O O O O O
   O O O O
    X X X
     X X
      X

Just for fun: задача с латвийской олимпиады. :)

Качественный контест! Спасибо!

Here is my submission to problem B: 2014265. It gives correct output on my system as well as ideone. But it fails on codeforces judge. Is there some problem with my code ? Comments are welcomed!!

My submission for Problem D results in runtime error at test case 35. i have no idea why this is happening. I am using the same concept as many other people during the contest.

I am getting an incorrect answer for problem B on test case #18. I wish there were a way to get the test case (it's being truncated currently).

I have found a very easy approach for problem A. The idea is to calculate like this if a>1 && b>1 && c>1 //it means its a hexagon: so ans = ans + 2*(a+b+c)-6 (calculating the tiles on perimeter)...

else //i.e. a==1 || b==1 || c==1 it means its a square now... ans = ans + (a*b*c);

for those who couldn't AC Problem D : i prefer learning function upper_bound(first_iterator, last_iterator, data); (in C/C++) (which uses binary_search and it's order is O(logN))

i think CLEAN BruteForce algorithm gets AC too... but with this function, implementing is SO EASY, it does all the job...

all you have to do is find four upper_bounds(two for each side of cell) and calculate distances see if they are equal...

In Problem B,

This Test Case :

6 4

1 2

1 3

4 5

4 6

The expected ans is 0 but how will we make 3-3 pairs with this configuration ?