Also about D:

If I am not wrong, then I think I have seen it on topcoder (in recent SRMs)

Edit: Problem D

in Output section:

be H if the i-th card should be placed with its front side visible, and T with its back.

actually ono should use duplicate volumes first

The following input should result in answer of 5, not 4.

7
1 2 4 1 6 7 271

try that case : 6 1 5 4 3 2 1

answer : 5

Counter test:

5
1 2 2 3 5

Actually, I just used binary search. It is easy to implement and not easy to get wrong. Here is the implementation, in case anyone is interested.

Counter example:

10
2 5 9 10 10 13 13 13 16 16

Expected: 6 Received: 5

Same here. Next time we'll do better!

I am assuming that you understood infinite sum part. Take a dp matrix of size 24*24, dp[i][j] = probability that next move is at jth hour if last move was at ith hour. The probability of 1st move at jth hour is dp[23][i], i.e, setting 23 as the last move. So we can get the 1*24 matrix which would store the probability of getting at ith hour after 1st move by multiplying matrix initialState(it is of size 1*24, all values except 23 is 0 and initialState[0][23] = 1) by dp matrix

Now let dp[i][j][k] (NOTICE: this is different state from editorial for simplification purpose first) = probability of ending at kth hour on ith move if last move was at jth hour. Now to get the probability of getting (i+1)th move on kth hour, let's assume that we are at xth hour on ith move and now we need to go on kth hour in next move, the conditional probability will be dp[1][x][k], so total probability will be dp[i][j][x]*dp[1][x][k]. So, dp[i+1][j][k] = $$$\sum_{x=0} ^{k-1} dp[i][j][x]*dp[1][x][k]$$$. So dp[i+1] can be written as dp[i]*dp[1], note that dp[i] and dp[1] are matrices of size 24*24

So dp[i] = $$$(dp[1])^{i}$$$

You can solve the above matrix with matrix exponentiation method and for getting probabilities you can multiply with matrix discussed in paragraph 1