Do not hope

You never gone blue

My hope has been granted :)

If you want, you can!

For the sake of fairness: azizkhan was one of the authors of Codeforces Round #202. We are just first, who prepared complete round.

Always dreamed to write this comment (

afarin be country shoma

you boro challenge24 bede we haminja contest midim

It's an official contest. Just appeared on the front page.

we too (:

miay teeworlds?

Ya, we know that by looking at the time you created your account at.

teach me how to spout nonsense without getting embarrassed please because the things you just said made laughing a medicine for me =))

I'm pretty sure everyone is so much concerned with their lives and won't dare to oppose you. But I was just wondering what if the Online Judge oppose you and give you a verdict other than Accepted!

I think he is just acting like the character Akashi Seijuro... Those words were all originally from the anime....

look for help bro :)

@Akashi: Sorry but you will loose both in the anime and in div1 soon :)

chi kos migi ?

Yes... of course that was first contest of ADJA we hope next contest be better :-"

People thought knight was K lololol

I use something like this: ....K... ........ ........ ........ ........ ........ ........ ........ (the code will fail this test case if they add 1 for 'K' or 'k')

mymap['K'] = mymap['k'] = 3; //this case

Also some people / not in my room :( / use 9 instead of 8 for Q.

I missed the contest but I guess it's related to LCA.

Preprocessing: LCA preprocessing and making an array of heights of nodes (h[]), quantities of nodes in the subtrees (sub[]), ancestors of levels 1,2,4,8,16... If we have ancestors of these levels, we can find ancestor of any level with O(log(N)) time.

Solution: if a==b, the answer is n. Else find LCA(a,b)=c. If heights of a and b are the same, the middle is c. We can find ancestors of a and b that are childs of c (call them e and f), because we know that they are a and b ancestors of the (h[a]-h[c]-1)th level (complexity O(log(N))). Then the answer is n-sub[e]-sub[f]. If heights of a and b are not the same, then either h[a]-h[b] is odd — then answer is 0, or is even — then we know where the middle node e is situated. If its child closer to a or b is f, the answer is sub[e]-sub[f].

I finally find my blade.

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If you know what I mean :-troll.

my solution is to print the minimum of (n+m)/3 and n and m

Greedily, whichever out of n or m is larger subtract 2 from it and subtract 1 from the other until one of them is smaller than 2 and the other smaller than 1.

Easiest solution i found was:

Ans=(a+b)/3

Ans=min(Ans,min(a,b))

I doubt this tho

ans=min(n,m,(n+m)/3) :)

i did something like this 1+max(solve(n-1,m-2),solve(n-2,m-1));where solve is my function.

i did this:

keep this : array[ each partial sum amount ][ each character ] = number of occurences of that character with the amount of partial sum.

(by partial sum i mean sum of all values from the first to the i'th character) since the amount of partial sum can be large and also negative you need map ( or unordered_map )

when you want to add a new partial sum you must see how many of the previous partial sum and the current character had occured and add this amount to the ans.

my submitted code is http://mirror.codeforces.com/contest/519/hacks/140490/test

you need to print "\n" at the end of your input.

You have a trailing space at the end of the lines

Use %d%c ... i == n - 1 ? '\n' : ' '

min(n,m,(n+m)/3) is not very challenging even 10^18. if this solution is correct...

Let's hope for fast rating updates too :D

UPD. It was very fast :) Thank you!

I don't understand it either. I submitted your code without any modifications and got AC 10086878

Try to write admins about that.

Ok)) I am need too))

LCA tutorial on topcoder.