lightoj 1030 Discovering Gold[ 期望 ]

B – Discovering Gold

Description

You are in a cave, a long cave! The cave can be represented by a 1 x N grid. Each cell of the cave can contain any amount of gold.

Initially you are in position 1. Now each turn you throw a perfect6 sided dice. If you get X in the dice after throwing, you addX to your position and collect all the gold from the new position. If your new position is outside the cave, then you keep throwing again until you get a suitable result. When you reach theNth position you stop your journey. Now you are given the information about the cave, you have to find out theexpected number of gold you can collect using the given procedure.

Input

Input starts with an integer T (≤ 100), denoting the number of test cases.

Each case contains a blank line and an integer N (1 ≤ N ≤ 100) denoting the dimension of the cave. The next line containsN space separated integers. The ith integer of this line denotes the amount of gold you will get if you come to theith cell. You may safely assume that all the given integers will be non-negative and no integer will be greater than1000.

Output

For each case, print the case number and the expected number of gold you will collect. Errors less than10-6 will be ignored.

Sample Input

3

1

101

2

10 3

3

3 6 9

Sample Output

Case 1: 101.0000000000

Case 2: 13.000

Case 3: 15

题目大意：从1位置摇骰子，到n位置，，每个位置有权值，求最后得到的权值期望；

代码：

姿势一：YY乱搞；

代码：

int v[110];double dp[110][110];double f[110];int main(){int T,cas=0;scanf("%d",&T);while(T–){int n;scanf("%d",&n);for(int i=1;i<=n;i++)scanf("%d",v+i);for(int i=0;i<=n;i++){f[i]=0.0;for(int j=0;j<=n;j++)dp[i][j]=0.0;}dp[1][1]=1.0;f[1]=1.0;int flag;for(int i=2;i<=n;i++){for(int j=1;j<=n-1;j++){flag=0;for(int k=6;k>=1;k–){if(!flag&&j+k<=n){flag=k;}if(flag){dp[i][j+k]+=dp[i-1][j]/flag;}}}for(int j=1;j<=n;j++)f[j]+=dp[i][j];}double ans=0.0;for(int i=1;i<=n;i++){// cout<<"-"<<f[i]<<endl;ans+=f[i]*v[i];}printf("Case %d: %.10lf\n",++cas,ans);}return 0;}姿势二：dp[i]表示重i到n得到权值的期望 dp[i]=v[i]+dp[i+k]/j;

代码：

int v[110];double dp[110];int main(){int T,cas=0;scanf("%d",&T);while(T–){int n;scanf("%d",&n);for(int i=1;i<=n;i++)scanf("%d",v+i);dp[n]=v[n];for(int i=n-1;i>=1;i–){int k=min(6,n-i);dp[i]=v[i];for(int j=1;j<=k;j++){dp[i]+=dp[i+j]/k;}}printf("Case %d: %.10f\n",++cas,dp[1]);}return 0;}

姿势三：dp[i] 经过i位置的概率，最后求ans;

代码：

int v[110];double dp[110];int main(){int T,cas=0;scanf("%d",&T);while(T–){int n;scanf("%d",&n);for(int i=1;i<=n;i++){dp[i]=0.0;scanf("%d",v+i);}dp[1]=1;for(int i=1;i<=n-1;i++){int k=min(6,n-i);for(int j=1;j<=k;j++)dp[i+j]+=dp[i]/k;}double ans=0.0;for(int i=1;i<=n;i++)ans+=dp[i]*v[i];printf("Case %d: %.10f\n",++cas,ans);}return 0;}

大把大把的时光从指缝间遛走，