Are you joking?

My question too...

Also in the contest page it's div1-div2 combined...

The post has been updated: now members of both divisions will participate in the same contest.

Then all div1 users would create div2 fakes

<3

I work with Diagram. Unfortunately, we aren't looking to offer internship opportunities to students outside of Canada at this time. It's fair game for Canadians though!

This is the first content we've sponsored but plan to do so again. This will definitely be a consideration in the future.

Read the third line of post

use your brain bro

It's here but apparently not in the front page yet.

UPD : It's on front page now.

i do it man

I work with Diagram. If you're already residing in Canada, just let us know in a "cover letter" via the application form or send me a note directly.

Условия будут доступны как на русском так и английском языках.

Someone posted blog entry http://codeforces.me/blog/entry/47858, which header includes malicious script

What does this code do <script>body.innerHTML="What does this code do?"</script>

So in "Recent actions" we have this script executed.

I work with Diagram. If any of the top 100 are local winners (Toronto or Montreal), we'll find some time to make this happen.

Congratulations you win the "Worst comment ever" prize keep up the good work.

no problem....they will start with 5 min delay as usual :p

Not that good of an idea right? Tried that a couple times, never works out...

I love them too , because in combined rounds you will never see unrated accounts in top 100

What is wrong?

No, just go on "Contests" page and click on "Register".

http://codeforces.me/contests/725

The leaderboard only shows members that took part in at least one contest in the last six months.

I think it is OK now.

Key abeservation is:

If you can make the greedy algorithm fail, you can do it by add just one coin. Prove it yourself. :)

So we can try all the C possibilities. And since there're at most unique numbers in the solution of greedy algorithms, you can use a fenwick tree to find the largest number which can be used in greedy algorithm. This method works in .

21687779

you can iterate all the value from 1 to C, consider it the value of the coin you give to Alfred, and with each value just do brute force, the time brute force running is no more than sqrt(C), so the total complexity is C * sqrt(C), i didn't realize the second part in the contest. My friend's code : http://ideone.com/56lb0F

yes

I think it's binary search

I used heap+sort! First of all, Let's sort ballons! Then, Let's insert everybody better,than us! Then, pop the people with minimum(weight-ballons+1)! Repeat it, until we can! Sorry for my english!

It's simple enough with scanf. scanf("%I64d%c", &x, &c);

It adds a bit of personality to the task if u ask me.

It is already simple. Isn't it?

long long row;
char column;
cin >> row >> column;

First time for everything man :D

Me too...

1. We find the repeating character.
2. Find two indices where this character is placed: i₁, i₂.
3. Find the number of characters between the repeating character: dist = i₂ - i₁ - 1.
4. If number of characters between them is 0 print "Impossible", otherwise it is possible and we proceed.

I look at the sequence in the following way:

So, basically, in the next steps I lay the loop (which is red in the picture) on the right side. The starting and ending characters I put on the left side (starting characters — on top and ending characters on bottom).

First calculate the distance between the two repeated letter. Then put the letters snake-shaped.

For example, if the distance is 5, then the rightmost part should be this:

   -> A -> x -> x
     /|         |
    / |         v
<- y  x <- x <- x

My submission here.

sorry for my bad English

There is the statement about this situation in the problem:

"If some copy of Alice receives messages from children nodes and also receives the answer she is waiting for at the same instant, then Alice first processes the answer, then immediately continue as normal with the incoming messages."

Sadness is the contest ends just before you click submit to an accepted solution :(

It's only impossible if there are two of the same letters next to each other. Also it's easily understandable that there will be exactly one letter that appears twice in the input string, since each letter must appear at least once and there's only one extra letter.

Let's say that this was the letter 'A'. Then the string looks like this:

s1 + "A" + s2 + "A" + s3

I got AC by constructing the matrix like this:

3333333A222

111111111222

It's always possible to get the input string by starting at the left start of the 1s, going right, taking 'A', getting the 2s by going right then down then left then back to 'A', then going left (and maybe down and right) to take all 3s.

Of course, s1 and s3 are of varied length and can even be empty, but then simply make the one that's too long "spill" into the row above or below. Also you have to be careful to write the string s1 in reverse order.

look at this comment

Not only you, no one can submit.

Not yet. It often only gets enabled after the ratings update.

I agree with you.

Tough luck

(I didn't AC it)

I think it is: always take the photo with the biggest sum of A and B. But the tricky part is to, at each point in time, check if it's possible that both players want to pass their turns, ending the game early. I'm pretty sure I've got the DP for this down, but I couldn't complete it during contest.

Yep. It can be done greedily if you regard the value of a card as a+b since the scores that the two players get are a,0 and 0,b in both cases, where the difference would be a and -b respectively. So you can set this value for each card and greedily select the pictures.

Of course you still have to handle some other cases :)

what a pity

I don't believe it.

did u use ideone or any other public online compiler?

I didn't actually read your code. Just tested it. time for 1st attendant and time for 2nd attendant should be the equal. For example you got 6e=25 so that 8e=25 but your output is 31.