One Person Game(機率+數學)

There is a very simple and interesting one-person game. You have 3 dice, namelyDie1, Die2 and Die3. Die1 has K₁ faces. Die2 has K₂ faces. Die3 has K₃ faces. All the dice are fair dice, so the probability of rolling each value, 1 to K₁, K₂, K₃ is exactly 1 / K₁, 1 / K₂ and 1 / K₃. You have a counter, and the game is played as follow:

1. Set the counter to 0 at first.
2. Roll the 3 dice simultaneously. If the up-facing number of Die1 is a, the up-facing number of Die2 is b and the up-facing number of Die3 is c, set the counter to 0. Otherwise, add the counter by the total value of the 3 up-facing numbers.
3. If the counter's number is still not greater than n, go to step 2. Otherwise the game is ended.

Calculate the expectation of the number of times that you cast dice before the end of the game.

 

Input

 

There are multiple test cases. The first line of input is an integer T (0 < T <= 300) indicating the number of test cases. Then T test cases follow. Each test case is a line contains 7 non-negative integers n, K₁, K₂, K₃, a, b, c (0 <= n <= 500, 1 < K₁, K₂, K₃ <= 6, 1 <= a <= K₁, 1 <= b <= K₂, 1 <= c <= K₃).

 

Output

 

For each test case, output the answer in a single line. A relative error of 1e-8 will be accepted.

 

Sample Input

2
0 2 2 2 1 1 1
0 6 6 6 1 1 1

Sample Output

1.142857142857143
1.004651162790698



題意: T 組數據, 每組數據一行，n, K1, K2, K3, a, b, c 表明 3 個骰子有 K1,K2,K3 個面
用這三個骰子玩遊戲，首先，計數器清零，擲一次，若是三個骰子中，第一個爲 a, 第二個爲b，第三個爲c ，計數器清零，不然，計數器累加三個骰子之和。
如此重複執行第二步 ，直到計數器和大於 n 問計數器大於 n 的遊戲次數指望

要推導出個遞推式子，而後發現都和 dp[0] 相關，分離係數，我也是看了這篇博客才懂的，寫得很好:
http://www.cnblogs.com/kuangbin/archive/2012/10/03/2710648.html

 1 #include <iostream>
 2 #include <stdio.h>
 3 #include <string.h>
 4 using namespace std;
 5 
 6 #define MAXN 600
 7 
 8 int n, k1, k2, k3, a, b, c;
 9 double A[MAXN],B[MAXN];
10 double p0;
11 double p[100];
12 
13 int main()
14 {
15     int T;
16     cin>>T;
17     while (T--)
18     {
19         scanf("%d%d%d%d%d%d%d",&n,&k1,&k2,&k3,&a,&b,&c);
20         memset(A,0,sizeof(A));
21         memset(B,0,sizeof(B));
22         memset(p,0,sizeof(p));
23 
24         p0=1.0/(k1*k2*k3);   //單位機率，變爲 0 的機率
25         for (int j1=1;j1<=k1;j1++)
26         for (int j2=1;j2<=k2;j2++)
27         for (int j3=1;j3<=k3;j3++)
28             if(j1!=a||j2!=b||j3!=c)
29             p[j1+j2+j3]+=p0;  //擲出某一個和的機率
30 
31         for (int i=n;i>=0;i--)//由於要循環到大於 n
32         {
33             for (int j=1;j<=k1+k2+k3;j++)
34             {
35                 A[i]+=p[j]*A[i+j];
36                 B[i]+=p[j]*B[i+j];
37             }
38             A[i]+=p0;
39             B[i]+=1.0;
40         }
41         double ans = B[0]/(1.0-A[0]);
42         printf("%.15lf\n",ans);
43     }
44     return 0;
45 }

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