lightOJ Discovering Gold(期望DP入门)

期望DP理解：http://kicd.blog.163.com/blog/static/126961911200910168335852/

kuangbin期望DP练习题：
http://www.cnblogs.com/kuangbin/archive/2012/10/02/2710606.html

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
Sample Output
Case 1: 101.0000000000
Case 2: 13.000
Case 3: 15

题目：从第一个格子出发每次可通过掷骰子往前条1~6格，并且收集到该格子的黄金，最终调到第n格。如果掷出的点数大与可以跳的格子，则继续掷骰子，直到掷出可以跳的格子。求最后调到n能够收集到黄金个数的数学期望。

对于期望的理解关键是定义式的理解。详见开头的链接。
这题 我们定义dp[i]表示从i跳到n的期望。
f[i]=f[i+1]/6+f[i+2]/6+f[i+3]/6+…+f[i+6]/6 +f[i];
为什么是这个式子呢？
可以手动模拟一下 泥会发现非常的神奇,这里的f[i]==num[i]
因为i这个点最终一定会跳到n 所以如果将按照定义式计算的i/p 分离出来 加在一起 f[i]=f[i]*1;
….好像说的好麻烦。。
举A B C D 四个数根据定义模拟一下就会明白~

#include <cstdio>#include <iostream>#include <cstring>#include <cmath>#include <algorithm>#define mem(array) memset(array,0,sizeof array)using namespace std;const int INF=1e8;double dp[105];int T,n;int main(){    freopen("in.txt","r",stdin);    scanf("%d",&T);    for(int cas=1;cas<=T;cas++){        scanf("%d",&n);        mem(dp);        for(int i=1;i<=n;i++)            scanf("%lf",&dp[i]);        for(int i=n-1;i>=1;i--){            double x=min(6,n-i);            for(int j=i+1;j<=i+x;j++)                dp[i]+=dp[j]/x;        }        printf("Case %d: %.7f\n",cas,dp[1]);    }    return 0;}