zoj-3329-期望/dp/方程优化

Posted 2020-11-02 zzq

One Person Game

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Time Limit: 1 Second      Memory Limit: 32768 KB      Special Judge

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There is a very simple and interesting one-person game. You have 3 dice, namely Die1, 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

 

 

　　　　f[i]表示已经得到i分之后距离目标的期望次数，pk表示得分为k的概率，则有f[i]=SUM{ p[k]*f[i+k] } + p0*f[0] + 1 ,因为下一次操作可能清零或者组合成其他点数，要分开讨论。这个式子无法直接递推，我们需要简化一下。

可以看出f[i]均和f[0]有关，不妨令f[i]=A[i]*f[0]+B[i] ,带入上式得到  f[i]=(sum{ p[k]*A[i+k] }  + p0)*f[0]+(SUM{ pk*B[i+k] } +1 ) ,可以看出A[i]=SUM{ pk*A[i+k] }+p0 , B[i]=SUM{ pk*B[i+k] }+1 ,A[i]和B[i]可以递推得到，所以答案就是A[0]/(1-B[0]);

　　　　

 1 #include<iostream>
 2 #include<cstring>
 3 #include<queue>
 4 #include<cstdio>
 5 #include<stack>
 6 #include<set>
 7 #include<map>
 8 #include<cmath>
 9 #include<ctime>
10 #include<time.h> 
11 #include<algorithm>
12 using namespace std;
13 #define mp make_pair
14 #define pb push_back
15 #define debug puts("debug")
16 #define LL long long 
17 #define pii pair<int,int>
18 #define eps 1e-12
19 
20 double p[50];
21 double A[1100],B[1100]; 
22 int main()
23 {
24     int n,m,i,j,k,t;
25     int k1,k2,k3,a,b,c;
26     scanf("%d",&t);
27     while(t--){
28         scanf("%d%d%d%d%d%d%d",&n,&k1,&k2,&k3,&a,&b,&c);
29         double p0=(double)1.0/k1/k2/k3;
30         memset(A,0,sizeof(A));
31         memset(B,0,sizeof(B));
32         memset(p,0,sizeof(p));
33         for(i=1;i<=k1;++i){
34             for(j=1;j<=k2;++j){
35                 for(k=1;k<=k3;++k){
36                     p[i+j+k]+=p0;
37                 }
38             }
39         }
40         p[a+b+c]-=p0;
41         for(i=n;i>=0;--i){
42             A[i]=p0;
43             B[i]=1;
44             for(j=0;j<50;++j)
45              A[i]+=p[j]*A[i+j],
46              B[i]+=p[j]*B[i+j];
47         }
48         printf("%.15f\n",B[0]/(1-A[0]));
49     }
50     return 0; 
51 }