LightOJ 1030 Discovering Gold (概率/期望DP)

题目链接：LightOJ - 1030

Description

You are in a cave, a long cave! The cave can be represented by a (1 imes 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

3

1
101

2
10 3

3
3 6 9

Sample Output

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

Solution

题意

有 (N) 个格子，每个格子有价值为 (val[i]) 的金子，初始你在第一个格子。

每次抛一个 (6) 面的骰子，抛到的数为 (X_i)，就往前走 (X_i) 个格子，如果超过格子 (N)，就重新抛，走到格子 (N) 就结束。求拿到金子的价值的期望。

思路

可以用 概率 DP 或 期望 DP 解决。

概率DP

求出到达每个格子的概率，然后乘上每个格子的价值再累加起来就行。

设 (dp[i]) 为到格子 (i) 的概率，则 (dp[i + j] = dp[i + j] + dp[i] / k (1le jle k, k = min(6, n - i)))。

期望DP

设 (dp[i]) 为格子 (i) 到 (N) 的能获得黄金的期望，则 (dp[N] = val[N])。

状态转移方程为 (dp[i] = frac{1}{k} sum_{j=1}^kdp[i+j] (k = min(6, n - i)))。

Code

概率DP

#include <bits/stdc++.h>
using namespace std;
const int maxn = 110;

int v[maxn];
double dp[maxn];

int main() {
    int T;
    scanf("%d", &T);
    int kase = 0;
    while(T--) {
        int n;
        scanf("%d", &n);
        for(int i = 1; i <= n; ++i) {
            scanf("%d", &v[i]);
        }
        memset(dp, 0, sizeof(dp));
        dp[1] = 1.0;
        for(int i = 1; i <= n; ++i) {
            int k = min(6, n - i);
            for(int j = 1; j <= k; ++j) {
                dp[i + j] += dp[i] * 1.0 / k;
            }
        }
        double ans = 0.0;
        for(int i = 1; i <= n; ++i) {
            ans += dp[i] * v[i];
        }
        printf("Case %d: %.7lf
", ++kase, ans);
    }
    return 0;
}

期望DP

#include <bits/stdc++.h>
using namespace std;
const int maxn = 110;

int v[maxn];
double dp[maxn];

int main() {
    int T;
    scanf("%d", &T);
    int kase = 0;
    while(T--) {
        int n;
        scanf("%d", &n);
        for(int i = 1; i <= n; ++i) {
            scanf("%d", &v[i]);
        }
        memset(dp, 0, sizeof(dp));
        dp[n] = v[n];
        for(int i = n - 1; i >= 1; --i) {
            dp[i] = v[i];
            int k = min(6, n - i);
            for(int j = 1; j <= k; ++j) {
                dp[i] += dp[i + j] * 1.0 / k;
            }
        }
        printf("Case %d: %.7lf
", ++kase, dp[1]);
    }
    return 0;
}