Hello all, my code gives correct answer for CASE 4 (Problem D) in local machine but here it fails. Can someone help me with this. Link for submission: code

I guess pretest 5 checks integer overflow.

still waiting :p

You can add the problem you solved in contest in practice problems :p

Hi, can you please explain why .

I can understand that E[ci] = E[v] + 1 but I could not understand the other part. Why are we taking the summation of number of vertices in subtree times the probability that Ci comes earlier than Cj. According to E[x] formula, shouldnt we do: E[c] = E[v] + 1 + (1/n!)*sigma(delay caused by i-th permutaion), where v is parent of c having n children.

Oh yes, thanks. Did not notice that :p

Why is the number of unique substrings after concatenating two strings is 100 +9?

Lets say two strings are there S1 and S2 of length 100. Both of them will have not more than 100 unique substrings of length 10. It is also possible that the 100 unique subtrings of S1 are not present in S2. So when we concatenate them we get a total of 100 (from S1) + 100(from S2) + 9(from border of two strings) unique substrings.

So if there are 100 such strings then the number of unique substrings can be: 100*100 + 9*100.

The wait is finally over :p

Never waited so long for the editorials.

Can someone please explain Leha and function — Adiv1 ?