suppose the best gift ever will be the invitation from a good company)))

You better create a blog post to invite to your channel instead of irrelevant comment here.

I just visited your channel. You should have mentioned that your videos are in Arabic language.

It always is, unless stated otherwise.

If you will get any help. Please share with me!

Participating in a programming contest is better than watching a 'fixed match'. I think.

Reloading the page is helpful

And me.

I have just installed a plug-in that has to be an alternative of adobe flash player, but no response.

Why it's not as the Educational rounds? I'm talking about the way of displaying the source code for hacking.

Please MikeMirzayanov, do something for this.

I used to face this problem on Google Chrome. Now I use Firefox for hacking, works all the time!

I stored an array of shifts for each level. When type 1 query comes, increase the shift for that level by K. When type 2 query comes, increase the shift for that level by K, and also the shift for the level after by 2k, and after that 4k... to the end of the array (which will be size 63 or something to get max number of elements). When type 3 query comes, find where that node is and go up by dividing by 2. Then you just need to subtract the shifts on each level as you go up. Sample: http://codeforces.me/contest/960/submission/37077181

If instead of keeping ai and bi, we keep ci= | ai — bi|, then our task is to use k1+k2 operations to reduce the cost. So you'll go through the array, find the maximum and decrease the value by 1. If all the elements are 0 it means that you are in an optimal point and the way to go from here is to add 1 and then decrease 1 to the same position up until you don't have any operations left.

Make a priority queue of the biggest differences between a and b. You can use the greedy approach then : just reduce 1 from the biggest difference that exist (i.e. remove biggest element from pq, reduce it by 1, and add the reduced number to the priority queue)

If the priority queue becomes empty while you still have attempts, the minimum result will just be number of remaining attempts modulo 2.

If you keep a sequence that is all 1's (ie 1 1 1 1 .. 1) the cost is 2^(no of 1's)-1. So you're going to find the largest k so that x>= 2^k -1 and then decrease x by 2^k -1 and add k equal values to your solution. You just have to be careful because the last k values should be larger than the others by at least D.

1000000000 1 is possible

268435356 1

1000000000 1

Centroid Decomposition

Yep. Just one improvised DFS will do the trick, and I would have done it, if not for the electricity going off...

the sequence of length n contains 2^n-1 sequences. Hence, the number X can be decomposed into powers of two minus 1. The degrees of the twos in this decomposition are the lengths of the sequences to be used. The degrees of the twos in this decomposition are the lengths of the sequences to be used. Collect on parts. The final sequence will look like:

1,1,1,....,1+(d+1),1+(d+1),....,1+(2*d+1),1+(2*d+1),...where the number of identical terms is equal to the powers of the twos to which we decomposed X.

If the number X cannot be completely decomposed into (2^i-1), then at the end of the answer it is necessary to add Another f distinct numbers, each of which will increase from the previous one by d+1.

Do not forget to take into account that the length of the answer should not exceed 10000

Consider case

2 100000

1 2 1

2 1 2

1 2 3

and so on. The simple dfs checks 50000 edges 50000 times, making it .

Surely it is better to put hard cases in the pretests than real tests... would you rather fail and not be able to fix it??

Holy shit, I added 2 * sum of the values to the answer. RIP good place

I found some pattern. Which included Pascal triangle and Stirling number of the first kind. I generated the required Stirling number with FFT but got WA :/ I think I should have used NTT or maybe there's a lot simpler solution or something.

I am very thankful that this did not change my rank! :P

I think it is large numbers or something, I got a run time error that I fixed by increasing the size of my array :P

There is a work around to do FFT under large Modulo. But in this problem it might be risky, as it is time consuming. The idea is -
Lets take C = 30000. Now partition each number A(x) as Ca_1i + a_2i, and B(x) as C_b1i + b_2i. Then convolute P = a₁ × b₁, Q = a₁ × b₂, R = a₂ × b₁, S = a₂ × b₂.
You can get the final result as (A × B)_i ≡ C²p_i + C(q_i + r_i) + s_i (mod M). It works because no number while doing fft corsses NC² and fits into precision. But you can't do fft directly into the main polynomial, then a number can be atmost NM², and that will overflow.

If you make a new graph so that the edges are the nodes and the nodes are the edges, you'll have exactly the same problem but you'll have to find the longest path of increasing nodes. This is done by going through the nodes in the order of their values and updating a dp[node]=longest increasing sequence ending in node.

It seems like dp and maintaining array of sets(couldn't implement it in time though).

Let's loop over the edges in reversed order and assume that the current edge will be the beginning of the path. This way we kinda eliminate the condition about indices being increasing and only have to worry about weights.

Define f(node)(w) as the maximum length of a path if we start at node with an edge of weight w. Let's say the current edge goes from A to B and has weight C. Now we have to update the value of f(A)(B) with the maximum value of f(B)(i) + 1 for i > C. Obviously keeping f(node) in the form of a segment tree, in which the indices are the weights of outgoing edges from node, does the job.

That's not how you read the arrays :P

ans should be long long imo

I don't understand that. "At least one 'a' and one 'b' exist" mean that there must at least one 'A' or one 'B' in the string. Then why not handling the case when there's no 'a' or 'b' will fail?