It is scored according to the difficulty

because even if someone all-killed ABC(before 212) before,he still can't get D(which seems exactly like what me did :(

Your implementtion is very complecated. I think simplest implementation is a simulation. Just try to find the min steps foreach single char. This works in O(n+m) since except the first char it allways works in 0 or 1 step. The only case we have to take care of is if there is no solution at all.

use BIT to get dp[cur] and change dp[lst] easily(which is pretty well explained in the editorial

code:My Code

We know that $$$y=(x-z)(x+z)$$$

Because $$$x-z\leq x+z$$$ , so we can just count the answer for every $$$x-z\leq 10^6$$$(Note that $$$x-z$$$ and $$$x+z$$$ have the same partity)

Lets consider $$$dp[i]$$$ to be the number of subsequences satisfying the conditions stated in the editorial and ending at index $$$i$$$. Now to calculate $$$dp[i]$$$, lets see where are all the possible second last elements could be ? It's easy to see that the second last element must be after or equal to the prev occurence of $$$A_i$$$. Let $$$j$$$ be the previous occurence of $$$A_i$$$. So essentially $$$dp[i] = \sum_{k=j}^{k=i-1}dp[k]$$$. However, notice that there might be multiple occurences of some element in the range $$$[j,i-1]$$$. In that case, we only need to add the dp value of the last occurence of that element. To take care of this we make sure that only the dp value of last occurence is active in the BIT while iterating through $$$i$$$ in increasing order.

In Conclusion you do the following at each $$$i$$$.

All these operations can be done using BIT/segment tree. Finally the answer would be

.

sorry guys, I think I get it