1030 Discovering Gold（水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
Output for Sample Input
3

1
101

2
10 3

3
3 6 9
Case 1: 101.0000000000
Case 2: 13.000
Case 3: 15
一行格子，每个格子上有一定数目的钱，你一开始站在位置1上，每次扔骰子，扔了几点走几步，但是扔出超过格子的点数是不算的，问这个过程中拿到的钱数的期望。

水概率dp，直接dfs记忆化搜索，因为走到这个格子，这个格子的期望是确定的，也就是当前（now）格子的期望和（next）期望关系如下
dp[now] = (a[next] + dp[next])/lea;
其中lea是能扔出的点数（因为点数过大有时候是作废的，因此每次概率不同），记忆化搜去吧。

#include<cstdio>
#include<cstring>
#include<iostream>
#include<queue>
#include<vector>
#include<algorithm>
#include<string>
#include<cmath>
#include<set>
#include<map>
#include<vector>
#include<stack>
#include<utility>
#include<sstream>
using namespace std;
typedef long long ll;
const int inf = 0x3f3f3f3f;
const int maxn = 1005;
int n,t;
int a[105];
double dp[105];
double dfs(int x)
{
    if(dp[x] >= 0)return dp[x];
    int lea = min(6,n - x);
    double ans = 0;
    for(int i = 1;i <= lea;i++)
    {
        ans = ans + (a[x + i] + dfs(x + i))/lea;
    }
    return dp[x] = ans;
}
int main()
{
    #ifdef LOCAL
    freopen("C:\\Users\\巍巍\\Desktop\\in.txt","r",stdin);
    //freopen("C:\\Users\\巍巍\\Desktop\\out.txt","w",stdout);
    #endif // LOCAL
    int kase = 1;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&n);
        for(int i = 0;i <= n;i++)dp[i] = -1;
        for(int i = 1;i <= n;i++)
            scanf("%d",a + i);
        printf("Case %d: %lf\n",kase++,a[1] + dfs(1));
    }
    return 0;
}