It seems to be lost?
T1
very strange expectations.
According to several conditions that can be found, the probability will eventually converge to the accuracy or less.
We just need to iterate enough rounds to.
T2
violence \ (O (n | S | 2 ^ n) \) have passed. . .
strange.
The answer is found to seek union, considered the foundation and cross the inclusion-exclusion.
Set \ (g (S), S \ subseteq A \) of \ (S \) in all cases in the (S \) \ number of programs in all common prefix string length * this happens.
\ (P (S) = \) , \ (S \) The number of question marks.
Then \ (g (S) \) contribution to global is \ (G (S) 2 ^ {P (the AS)} \)
\ [ANS = \ SUM \ limits_ {S \ subseteq A} (-. 1) ^ { | S | -1} G (S) ^ 2 {P (the AS)} \]
\ (Hang [I] [J] \) represents the \ (I \) a question mark and a prefix strings.
\ (DP [I] \) , \ (S \) common prefix length of all strings \ (I \) , without considering \ (I \) scheme under the influence of the subsequent portion of the case.
\ [gs [j] = \
sum \ limits_ {i \ in S} hang [i] [j] \] wherein \ (J \ in [. 1, min \ len {\}] \) .
If \ (S \) of each string \ (I \) bit is not\ (1 \) , then the \ (CH [I] [0] = 1 \) , \ (CH [I] [1] \) is similarly defined.
\ (dp [i] = dp
[i-1] * [ch [i] [0] == 1 || ch [i] [1] == 1] \) Another full transfer of a question mark \ ( GS [I +. 1] -gs [I] == | S | \ rightarrow DP [I] = 2 * \)
\ [G [S] = \ SUM \ limits_. 1 {I} = min {^ \ len {\ }} dp [i] * 2
^ {gs [max \ {len \}] - gs [i]} \] the reason for this is to consider the contribution made by each position.
So you can find the answer.
Pretreatment what can be done \ (O (| S | 2 ^ n) \)
T3
without first fraction of points.
Guess conclusion.
If found except \ (i \) radar outside of all determine the height, then \ (i \) of optimal decision must be \ (0 \) or \ (y_i \) .
Consider a \ (O (n ^ 2) \) violence.
According to \ (x + y \) sorted then direct violence \ (DP \) .
In fact the enumeration within the range of 300 to the card can \ (AC \) a.
Metastatic according to \ (L_i \) divided into two parts.
For: \ (r_j <= L_i \) can therefore be directly prefix \ (max \) + half.
If: \ (r_j> L_i \) .
This part of our decision-making requires something monotonic.
Commentary title is engaged in a monotonous half stack on the line.
Then listen to face brother talk a bit and found not care if \ (log ^ 2 \) the complexity of the case, in fact, directly \ (cdq \) + decision monotony template on it.
First update interval with the left and then the right side to the left in the current update, and then deal with the right on the line.