Discovering Gold LightOJ - 1030 （概率期望）

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这是一道关于概率期望的数学问题，描述了在洞穴中寻找黄金的过程。玩家在1到N的网格中移动，掷骰子决定前进步数并收集相应位置的黄金。当到达N号位置时停止。给定每个位置的黄金数量，你需要找出能收集到的黄金的期望值。题目提供了输入输出示例，并指出可以通过动态规划从后往前计算期望值来解决。

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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

大致题意：n个格子，编号1到n，每个格子里面有一些金币，然后你有一个1~6的色子，甩到哪个数就往前走几步，然后取得到达的格子里的金币，1~6的几率相同，最后达到n号格子或者走出这n个格子时结束。问最后获得的金币的期望。

思路：期望dp，从后往前推

代码如下

#include<iostream>
#include<set>
#include<vector>
#include<algorithm>
#include<cstring>
#include<cstdio>
using namespace std;
typedef long long ll;
int a[105];
double dp[105]; 
int main()
{
    int T;
    int n;
    scanf("%d",&T);
    for(int cas=1;cas<=T;cas++)
    {
        scanf("%d",&n);
        for(int i=1;i<=n;i++)
        {
            scanf("%d",&a[i]);  
            dp[i]=a[i];
        }
        for(int i=n;i>=1;i--)
        {
            int step=min(6,n-i);
            for(int j=1;j<=step;j++)
            dp[i]+=1.0/step*dp[i+j];
        }
        printf("Case %d: %.7lf\n",cas,dp[1]);
    }
    return 0;
}