LightOj-1030 Discovering Gold (期望DP)

最新推荐文章于 2021-10-20 17:33:08 发布

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最新推荐文章于 2021-10-20 17:33:08 发布

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原文链接：http://www.cnblogs.com/songorz/p/9483740.html

本文探讨了一种独特的寻宝游戏算法，在一个长洞穴中，玩家通过掷六面骰子前进并收集黄金，目标是计算达到洞穴终点时预期收集的黄金数量。文章提供了一个详细的解决方案，包括代码实现，展示了如何使用动态规划来解决这个问题。

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 N^th 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 i^th integer of this line denotes the amount of gold you will get if you come to the i^th 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

101

10 3

3 6 9

Sample Output

Case 1: 101.0000000000

Case 2: 13.000

Case 3: 15

题解：

起始位置是1，从1走到n，给你一个骰子（6个面），按点数走，收集每一点上的金子，如果你将要走到的位置在n之内，就继续扔，往前走，如果在n之外，就一直扔到合适的位置为止，求到达n点时的期望

这个题是一个求期望的题，那么值得注意的是，当扔在n之外的情况是无效的，所以我们在位置i<n−6

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 int T,n;
 4 double dp[110];
 5 
 6 int main()
 7 {
 8     scanf("%d",&T);
 9     for(int k=1;k<=T;k++)
10     {
11         scanf("%d",&n);
12         memset(dp,0,sizeof dp);
13         for(int i=1;i<=n;i++) scanf("%lf",dp+i);
14         for(int i=n-1;i>=1;i--)
15         {
16             for(int j=1;j<=6;j++) dp[i]+=dp[i+j]/(1.0*min(6,n-i));
17         }
18         printf("Case %d: %.10lf\n",k,dp[1]);
19     }
20     
21     return 0;
22 }