CF235 Let & amp; #39; s Play Osu! [Dp + probability]

CF235 Let & #39; 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

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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

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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.

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     #include using namespace std;const int NN=111111;double f[NN];double dp[NN];int main(){#ifndef ONLINE_JUDGEfreopen("/home/rainto96/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);}