B - Discovering Gold（概率期望dp）

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 perfect 6 sided dice. If you get X in the dice after throwing, you add X 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 the Nth position you stop your journey. Now you are given the information about the cave, you have to find out the expected 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 contains N space separated integers. The ith integer of this line denotes the amount of gold you will get if you come to the ith cell. You may safely assume that all the given integers will be non-negative and no integer will be greater than 1000.

Output
For each case, print the case number and the expected number of gold you will collect. Errors less than 10-6 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
一维迷宫
题意：从位置1开始，n是迷宫深度，投骰子数xj，就走当前位置+x（走出迷宫则重新投），直到走到第n个位置为止。
由于每次不一定走几步，存在概率问题，求在各种情况下最后的期望
走出迷宫则重新投：这条使得概率在保持分子是1（例如剩余两步到终点，概率就是1/2，1/2而不是1/6，5/6）

//dp[]：所在位置经掷骰子到终点所获得的的金币的期望
// 即所在位置到终点 我们要求的 所以dp[1]为最终结果从起点到终点

#include 
int main()
{
    int i,j,k,T,n,a[1001];
    double dp[1001];
    scanf("%d",&T);
    for(k=1;k<=T;k++)
    {
        scanf("%d",&n);
        for(i=1;i<=n;i++)
        {
            scanf("%d",&a[i]);
        }
        dp[n]=a[n];
        double t,sum;
        for(i=n-1;i>=1;i--) //由后往前 填表
        {
            t=n-i;//由距离终点的距离判断概率 
            if(t>6) t=6;//骰子六个数使得概率不能小于 1/6
            sum=0;
            for(j=i+1;j<=i+t;j++)
            {
                sum+=(a[i]+dp[j])*1.0/t;//   1.0/t是概率
            }
            dp[i]=sum;
        }
        printf("Case %d: %lf\n",k,dp[1]);
    }
    return 0;
}