Of course yes.

only div2 users

Usually, in contests with ACM-ICPC rules problems don't sorted according to difficulties.

тот который справа — видимо до конца регистрации, слева — до соревнования.

На французском) А ещё, для тех , кому необходимо, будут так же доступны условия на китайском)

На русском и английском.

I have this problem too, Can you solve it? Also there are different time to contest beginning, what does it mean?

Попробуйте сдать какую-либо простую задачу на PHP в архив задач. Вот пример решения задачи 233A - Идеальная перестановка

<?php
$a = range(1, trim(fgets(STDIN)));
if(count($a) % 2 == 0)
{
    $b = implode(" ", array_reverse($a));
    print $b;
}
else
{
    print "-1";
}
?>

Contest duration is 2 hours.. Wait for some time , you will know the number of problems

They would not do it as there are no rooms !

No more than 10 minutes. About 8-9 minutes mostly. We are sorry about the queue, it is the result of ACM-ICPC mode + very easy tasks for many participants. We will make conclusions and prevent this next time. Sorry again about 9 minutes queue.

Happy that contest is held on CF , upset that CF has a bit weak server. :|

unrate because your result is unsatisfactory? conditions were the same for all of the participants, am I right?

Нет, мы убрали те тесты, на которых ничего не упало на момент их удаления и которые выглядели однотипно с существующими. К сожалению, заранее понять какие именно тесто окажутся ключевыми невозможно — иногда решения падают на тестах, которые мало отличаются от предыдущих, но тем не менее решения падают.

IN EDIT

just count it from behind.. if n=7 then count how much we take coin from - 3 6 7 (we must make chest-6 and chest-7 empty) - 2 4 5 (we must make chest-4 and chest-5 empty) - 1 2 3 (we must make chest-2 and chest-3 empty)

then add the coin left in chest-1

Agree, i think many other people will agree too.

I am also waiting for the same.

So the Rating is in queue too.

updated. too fast today (as no system test)

from any record before the current record

wrong testcases? didn't notice..

It's not. Your input is wrong, since it is impossible to generate it in the way, described in statement.

and solutions which accepted input 3 -1 1000000000 1 1000000000 -1 1 1 1 -1 out put 1000000001 1000000001 1 and problem (D) said (0 ≤ ai ≤ 109) CONTRADICTION! how to accept it?

И увы, значение опыта огромное. Благодаря ему программист не только меньше допускает глупых ошибок, но и гораздо быстрее придумывает решение и реализовывает его. Ведь, имея за плечами тысячу решенных задач, легче определить тип задачи и думать в правильном направлении.

Yes. dp[i][j] = number of palindromes in the [i,j] substring.

First you check this:

isp[i][j] = "is the substring [i,j] palindrome?" After that you can try this simple DP:

dp[i][j] = number of palindromes in the [i,j] substring.

So it's easy to see that dp[i][j] is equal to dp[i+1][j] + dp[i][j-1] — dp[i+1][j-1] + isp[i][j].

The intersection between [i,j-1] and [i+1,j] is equal to [i+1,j-1], so "dp[i+1][j] + dp[i][j-1]" calculates dp[i+1,j-1] twice and you've to subtract this one more time to maintain the correct result. Also, you've to add 1 in the solution if the substring [i,j] is palindrome.

A lot of contestants solved this problem by using Manacher Algorithm + DP, but it's only a matter of time complexity.

Queue was formed based on submission times. Those that solved first two problems really fast were in the beginning of the queue.

It is obvious thing.. The queue was only formed becuase there were many submissions which the judge could not handle all at once. Obviously during the first 5 minutes there were very few people who made submissions and so received quick judgements. And no points for guessing that these were high rated coders.

I hope you had done a lot of research before posting this , becuase this is a algorithmic platform , judge here is not a rascist and decisions are not taken here based on colors .**Colors only affect when you get upvotes and downvotes in the community** :P

The idea is similar to the last CF problem D (div 2). There are N guys (N is at most 2*M) and M pair of friends. You just do a brute force to check for all x, how many y has the most common friends (to check common friends you need a linear algorithm, like sorting everyones friends and doing a two pointers thing, or even easier calling set_intersection ;) ). It seems to be cubic, but its not, for one guy you do at most N*A + M operations (A is the number of his friends), the total complexity will be N*A + M + N*B + M + N*C + M + ... = N*(A+B+C..) + N*M = N*M + N*M = O(M^2). Here is my implementation if it is not clear yet: http://codeforces.net/contest/245/submission/2602015

I hope it helps :)

А минусующие, видать, знают ответы на мои вопросы, но набирать текст они не умеют, вместо этого минус нажимают.