Cf235 let's play Osu! [DP + probability]

Question:

For N positions, the probability pI of O occurrence at each position on 1-N is given. The scoring rules are as follows. x consecutive OSS records are x ^ 2 points, and the sum is calculated. For example, xxoooxooxx scores are as follows:

Expected score

Think about it. We can easily obtain the O (N ^ 2) method.

Make DP [I] The expected score of the former I

So

Obviously, this question

Consider how to change the score

We have

Then the scoring method becomes

Number of consecutive o pairs × 2 + O

One o can contribute 2 points

Now the score source is changed to two places

A pair of O (2 points) and a single O (1 point)

We know

Expectation = probability x benefit

We find the probability of each pair of O × 2

Find the probability of occurrence of a single O X

Sum is the expected score

For a certain vertex I

There are (1, I) (2, I) (3, I) (4, I )... (I-1, I) These pairs of O, each probability is from left to right concatenation, such

Make these probabilities equal to DP [I], that is

In this way, we have a recursive relationship.

For I + 1,

The sum of DP sum is the sum of all probabilities of occurrence of o x 2

+

Probability of a single o x 1

The expected score is obtained.

 
# Include <cstdio> # include <cstring> # include <cmath> # include <iostream> # include <algorithm> using namespace STD; const int nn = 111111; double f [NN]; double DP [NN]; int main () {# ifndef online_judgefreopen ("/home/raw.96/in.txt", "r", stdin ); # endifint N; scanf ("% d", & N); double sum = 0; For (INT I = 1; I <= N; I ++) {// CIN> F [I]; scanf ("% lf", & F [I]); sum + = f [I];} double ansum = 0; for (INT I = 2; I <= N; I ++) {DP [I] = (DP [I-1] + F [I-1]) * f [I]; ansum + = DP [I];} printf ("% F \ n", ansum * 2.0 + sum );}
 

Cf235 let's play Osu! [DP + probability]