Yes

I actually forget every time what it means. It means 700+200 or 500+200?

There has always been English editorials for all Atcoder Grand Contests.

Tasks weren't that hard in my opinion, at least implementation was easy. They required more thinking instead of coding.

It is because of all the silly people like me who do not know how to use bitmask dp. (for example, IOI p2 subtask 2)

EDIT: Oops. Since you solved E I assumed you meant it was easy.

Is there any reason why you don't make these rounds part of atcoder regular/grand contests?

I did (well upsolved, I couldn't participate in the contest). I know your question is pretty old but I thought that you might find the answer useful in some way. I just kept some sort of dp: lst[i][j] = the id of the last operation that affected the node j in such a way that j, at its turn, would affect the nodes at distance i from itself. The recurrence is something like update for each (i, u), lst[i — 1][v] for each v such that v is a neighbour of u with lst[i][u]. You could check out my source for more details: http://agc012.contest.atcoder.jp/submissions/1452611

Thanks for the very nice problem set. I tried to solve F, understood the part in the editorial saying "there is no (i, j) pair with i < j, and b_j < b_i < b_j + 1", but I am stuck at how to go from there to the dynamic programming solution dp[i][j][k], where i is the number of values filled from the end, j is the number of candidates, and the k-th candidate is chosen. Could someone elaborate on this part ?

It is the critical point that a₁'s color is different from a₂'s color.

Basically, we can use a₁ to change the position of balls. However we have to use a₂ to change the position of balls which color are the same of a₁. We don't need to use the other balls in operation2. Because the other balls' weight are heavyier than a₁ and a₂.

Could you got it?

Here is an alternative solution, which also explains that:

1. For each color, get the minimum value and replace every weight with minimum if min + original_weight <= X (it means that you can replace the balls within the same color).
2. Sort such pairs (new_weight, color).
3. Generally you should be able to permute all the balls such that: new_weight_i + min_new_weight <= Y. The rest stays the same. The only exception are such balls where color_i == color_of_min_new_weight, which can't be replaced because sum of their weights > X. The only way to move such a ball is to use the smallest new_weight with a different color. That's why you need second_min_new_weight and if sum of their weights <= Y, you can move such a ball, otherwise it must stay at original position.
4. You get all movable balls of size n, you count n! and divide it by factorial of the count of the colors included in movable balls set: n!/(k1!k2!..ki!)

When I started to think this problem, the definition of good string is different from current definition. The condition is like this: x can be represented as yyzz (not yy).

Under this constraints, it is tough to use a character three times. Thus, I came up with this approach.