One Person Game(几率+数学)

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



题意: T 组数据, 每组数据一行，n, K1, K2, K3, a, b, c 表明 3 个骰子有 K1,K2,K3 个面
用这三个骰子玩游戏，首先，计数器清零，掷一次，若是三个骰子中，第一个为 a, 第二个为b，第三个为c ，计数器清零，不然，计数器累加三个骰子之和。
如此重复执行第二步 ，直到计数器和大于 n 问计数器大于 n 的游戏次数指望

要推导出个递推式子，而后发现都和 dp[0] 相关，分离系数，我也是看了这篇博客才懂的，写得很好:
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);   //单位几率，变为 0 的几率
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;  //掷出某一个和的几率
30 
31         for (int i=n;i>=0;i--)//由于要循环到大于 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 }

View Code