題干：

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

題目大意：

在一個1＊N的格子里，每個格子都有相應的金幣數(shù)，走到相應格子的話，就會得到該格子的金幣。?
現(xiàn)在有一個人在1這個位置，手里有一顆色子，色子搖到幾，他就前進幾步，但有一種情況例外，如果當前位置＋色子數(shù) > N,那么他就會重新?lián)u色子。?
走到N位置，意味著游戲結束了。?
問游戲結束時，這個人得到金幣的期望。

解題報告：

首先根據(jù)期望的線性性轉化題意：考慮每個點被選中的概率，走到每個點的概率乘以該點的權值 的和。

dp[i]表示走到i這個點的概率是多少?！拔覟槿巳恕狈ㄞD移。（最后的dp[n]肯定==1）

最后乘以權值就是答案。

AC代碼：

#include<cstdio> #include<iostream> #include<algorithm> #include<queue> #include<map> #include<vector> #include<set> #include<string> #include<cmath> #include<cstring> #define F first #define S second #define ll long long #define pb push_back #define pm make_pair using namespace std; typedef pair<int,int> PII; const int MAX = 2e5 + 5; int n,a[MAX]; double dp[MAX]; int main() {int t,iCase=0;cin>>t;while(t--) {scanf("%d",&n);for(int i = 1; i<=n; i++) scanf("%d",a+i);printf("Case %d: ",++iCase);for(int i = 1; i<=n+6; i++) dp[i]=0;dp[1]=1;for(int i = 1; i<=n; i++) { // for(int j = 1; j<=6; j++) { // if(i-j >= 1) dp[i] += dp[i-j]; // }int up = min(n-i,6); for(int j = 1; j<=up; j++) dp[i+j] += dp[i]/up;}double ans = 0;for(int i = 1; i<=n; i++) ans += dp[i] * a[i];printf("%.8f\n",ans);}return 0 ; }

思路2：（鏈接）

這題要倒著推，由N推向1?
設dp[k]為到達k這個位置時得到金幣的期望，m為該點和N這個位置的距離，gold[k]為k這個位置的金幣數(shù)，因為走的位置不能超過N，所以要取min(m,6)?
那么dp[k] = 1 / min(m,6) * (dp[k + 1] + dp[k+2] + … + dp[min(m,6)]) ＋ gold[k]。

（其實是因為這題巧了，所以dp[k]可以等概率的由后面的點轉移而來，但是第一種方法更具有普遍性）

#include<cstdio> #include<cmath> #include<algorithm> using namespace std; using namespace std; const int N = 110; int gold[N]; double dp[N]; int n;double solve() {for(int i = 0; i < n - 1; i++) dp[i] = 0;int step;dp[n-1] = gold[n-1];for(int i = n - 2; i >= 0; i--) {dp[i] = gold[i];step = min(6, n-1-i);for(int j = 1; j <= step; j++)dp[i] += 1.0 / step * dp[i+j];}return dp[0]; }int main() {int test, cas = 1;scanf("%d", &test);while(test--) {scanf("%d", &n);for(int i = 0; i < n; i++)scanf("%d", &gold[i]);printf("Case %d: %.7lf\n", cas++, solve());}return 0; }

?

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