LightOJ1030---Discovering Gold(概率dp)

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

Output for Sample Input

3

1

101

2

10 3

3

3 6 9

Case 1: 101.0000000000

Case 2: 13.000

Case 3: 15

Problem Setter: Jane Alam Jan

dp[i]表示在第i个格子时得到的金子的期望,然后记忆化搜索一下就行了

/*************************************************************************
    > File Name: b.cpp
    > Author: ALex
    > Mail: [email protected] 
    > Created Time: 2015年04月29日 星期三 19时04分23秒
 ************************************************************************/

#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 

using namespace std;

const double pi = acos(-1.0);
const int inf = 0x3f3f3f3f;
const double eps = 1e-15;
typedef long long LL;
typedef pair <int, int> PLL;

double dp[110];
int arr[110];
int n;

double dfs(int cur) {
    if (dp[cur] != -1.0) {
        return dp[cur];
    }
    double ans = 0;
    int cnt = 0;
    for (int i = 1; i <= 6; ++i) {
        if (cur + i <= n) {
            ++cnt;
        }
    }
    for (int i = 1; i <= 6;  ++i) {
        if (cur + i <= n) {
            ans += (1.0 / cnt) * dfs(cur + i);
        }
    }
    ans += arr[cur];
    dp[cur] = ans;
    return dp[cur];
}

int main() {
    int t;
    scanf("%d", &t);
    int icase = 1;
    while (t--){
        scanf("%d", &n);
        for (int i = 1; i <= n; ++i) {
            scanf("%d", &arr[i]);
            dp[i] = -1.0;
        }
        dp[n] = arr[n];
        printf("Case %d: %.12f\n", icase++, dfs(1));
    }
    return 0;
}