传送门:http://www.lightoj.com/volume_showproblem.php?problem=1030
Discovering Gold
Time Limit: 2 second(s) Memory Limit: 32 MB
Program 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 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
Output for Sample Input Case 1: 101.0000000000
Case 2: 13.000
Case 3: 15
【动态规划-概率dp|LightOj（1030-Discovering Gold（期望dp模板））】解题心得：
- 考的就是一个期望dp ,求期望一个很重要的就是逆求期望，为啥是逆求呢，如果正求是在前面期望的基础之上求期望。期望的期望只是一种可能性，并不符合概率要求。这个可以参考贝叶斯公式的定义，里面说的很清楚。关于求期望，要从已知推到未知，就这个题来说，已知只能是必定到达最后一个格子。所以要从已知走向位置就是逆着求的。
- dp[i]代表的是扔到第i个格子期望得到多少的金子。
#include
using namespace std;
const int maxn = 110;
double dp[maxn];
int num[maxn],t,n;
int main()
{
scanf("%d",&t);
int T = 1;
while(t--)
{
int n;
scanf("%d",&n);
for(int i=1;
i<=n;
i++)
{
scanf("%d",&num[i]);
dp[i] = num[i];
}
int count = 1;
for(int i=n-1;
i>=1;
i--)
{
if(count > 6)//在后面没有六个格子可能性就没有六种
count = 6;
for(int j=1;
j<=count;
j++)
dp[i] += (double)(dp[i+j]/count);
count++;
}
printf("Case %d: ",T++);
printf("%.7f\n",dp[1]);
}
} 
