lightoj 1030 Discovering Gold (基礎概率dp)

_TCgogogo_發表於2015-09-24
Go

Discovering Gold
Time Limit: 2000MS   Memory Limit: 32768KB   64bit IO Format: %lld & %llu


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


題目連結：http://lightoj.com/volume_showproblem.php?problem=1030


題目大意：一條線上n個點，每個點有不同數量的金子，每次投骰子1-6，投到幾前進幾步並得到對應點上的金子，如果點數超過n則繼續投，直到到n為止，求得到金子數量的期望值


題目分析：很基礎的概率dp，設dp[i]為從i到n所能得到的金子的期望，顯然dp[n] = val[n]，然後從後往前遞推，dp[i - 1] += Σdp[i] / 6，但是注意有可能加6會超過n，所以遞推式應該為dp[i - 1] += Σj (1-6) dp[i] / min(6, n - i)

#include <cstdio>
#include <algorithm>
using namespace std;
int const MAX = 105;
double dp[MAX];

int main()
{
    int T;
    scanf("%d", &T);
    for(int ca = 1; ca <= T; ca++)
    {
        int n;
        scanf("%d", &n);
        for(int i = 1; i <= n; i++)
            scanf("%lf", &dp[i]);
        for(int i = n - 1; i >= 1; i--)
            for(int j = 1; j <= 6; j++)
                dp[i] += dp[i + j] / (1.0 * min(6, n - i));
        printf("Case %d: %.8f\n", ca, dp[1]);
    }
}


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