Acdream 1113 The Arrow (probability dp)

Acdream 1113 The Arrow (probability dp)

 

Question:

The initial state is 0. Each time a dice is dropped [], if it is in x, the number of dice is y, and if x + y> n, it is still in x.

Calculate the expected number of times the dice are dropped from 0 to n.

Analysis:

This is changed from the previous one: if the number of dice is y in x, and if the number of dice is y in x + y> n, the number of dice remains in x.

Then we set dp [I] to indicate the expected number of dice to be dropped from I to n.

So

We set the possibility to stay in the same place every time x times.

Dp [I] = dp [I] * y/6 + dp [I + 1]/6 + dp [I + 2]/6... + 1;

Then the formula of dp [I] is simplified.

The Code is as follows:

#include 
 
  #include 
  
   #include 
   
    #include using namespace std;const int maxn = 1e5+10;double dp[maxn];int main(){    int t,n;    scanf(%d,&t);    while(t--){        scanf(%d,&n);        memset(dp,0,sizeof(dp));        for(int i=n-1;i>=0;i--){            int tot=0;            double tmp=0;            for(int j=1;j<=6;j++){                if(i+j>n) tot++;                else tmp+=dp[i+j]/6.0;            }            //cout<