正推，到达i的概率为p[i]，要注意除了1和n外，到达i的概率并不一定为1
概率表达式为p[i] += p[j] / min(n - j, 6)
从j带过来的期望为exp[i] += exp[j] / min(n - j, 6)
又到达i时有价值val[i]，到达i的概率为p[i],故exp[i] += val[i] * p[i]
#include<cstdio>
#include<iostream>
#include<cstdlib>
#include<cstring>
#include<string>
#include<algorithm>
#include<map>
#include<queue>
#include<vector>
#include<utility>
using namespace std;
typedef long long LL;
const int N = 1008, INF = 0x3F3F3F3F;
#define MS(a, num) memset(a, num, sizeof(a))
#define PB(A) push_back(A)
#define FOR(i, n) for(int i = 0; i < n; i++)
double exp[N], p[N];
int val[N];
int main(){
int t;
cin>>t;
for(int cas= 1; cas <= t;cas++){
int n;
cin>>n;
for(int i =1; i <= n; i++){
scanf("%d", &val[i]);
}
p[1] = 1;
exp[1] = val[1];
for(int i = 2; i <= n; i++){
exp[i] = p[i] = 0;
for(int j = max(1, i - 6); j < i; j++){
exp[i] += exp[j] / min(n - j, 6);
p[i] += p[j] / min(n - j, 6);
}
exp[i] += val[i] * p[i];
}
printf("Case %d: %.10f
", cas, exp[n]);
}
return 0;
}