zoj3329 概率dp求期望

http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=3754

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:
Set the counter to 0 at first.
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.
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

/**zoj3329
dp求期望的题。
题意：
有三个均匀的骰子，分别有k1,k2,k3个面，初始分数是0，
当掷三个骰子的点数分别为a,b,c的时候，分数清零，否则分数加上三个骰子的点数和，
当分数>n的时候结束。求需要掷骰子的次数的期望。
题解:
设 E[i]表示现在分数为i，到结束游戏所要掷骰子的次数的期望值。
显然 E[>n] = 0; E[0]即为所求答案;
E[i] = ∑Pk*E[i+k] + P0*E[0] + 1; (Pk表示点数和为k的概率，P0表示分数清零的概率)
由上式发现每个 E[i]都包含 E[0]，而 E[0]又是我们要求的，是个定值。
设 E[i] = a[i]*E[0] + b[i];
将其带入上面的式子：
E[i] = ( ∑Pk*a[i+k] + P0 )*E[0] + ∑Pk*b[i+k] + 1;
显然，
a[i] = ∑Pk*a[i+k] + P0;
b[i] = ∑Pk*b[i+k] + 1;
当 i > n 时：
E[i] = a[i]*E[0] + b[i] = 0;
所以 a[i>n] = b[i>n] = 0;
可依次算出 a
,b
; a[n-1],b[n-1] ... a[0],b[0];
则 E[0] = b[0]/(1 - a[0]);
*************************************************************************
总结下这类概率DP:
既DP[i]可能由DP[i+k]和DP[i+j]需要求的比如DP[0]决定
相当于概率一直递推下去会回到原点
比如
(1):DP[i]=a*DP[i+k]+b*DP[0]+d*DP[i+j]+c;
但是DP[i+k]和DP[0]都是未知
这时候根据DP[i]的方程式假设一个方程式：
比如:
(2):DP[i]=A[i]*DP[i+k]+B[i]*DP[0]+C[i];
因为要求DP[0],所以当i=0的时候但是A[0],B[0],C[0]未知
对比(1)和(2)的差别
这时候对比(1)和(2)发现两者之间的差别在于DP[i+j]
所以根据(2)求DP[i+j]然后代入(1)消除然后对比(2)就可以得到A[i],B[i],C[i]
然后视具体情况根据A[i],B[i],C[i]求得A[0],B[0],C[0]继而求DP[0]
请看这题：http://acm.hdu.edu.cn/showproblem.php?pid=4035
*************************************************************************
**/

#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
using namespace std;

double a[505],b[505],p[505];
int n,k1,k2,k3,x,y,z;

int main()
{
int T;
scanf("%d",&T);
while(T--)
{
scanf("%d%d%d%d%d%d%d",&n,&k1,&k2,&k3,&x,&y,&z);
double p0=1.0/(k1*k2*k3);
memset(p,0,sizeof(p));
for(int i=1; i<=k1; i++)
{
for(int j=1; j<=k2; j++)
{
for(int k=1; k<=k3; k++)
{
if(i==x&&j==y&&k==z)
continue;
p[i+j+k]+=p0;
}
}
}
memset(a,0,sizeof(a));
memset(b,0,sizeof(b));
for(int i=n; i>=0; i--)
{
for(int j=3; j<=k1+k2+k3; j++)
{
a[i]+=a[i+j]*p[j];
b[i]+=b[i+j]*p[j];
}
a[i] += p0;
b[i] += 1;
}
printf("%.15lf\n",b[0]/(1-a[0]));
}
return 0;
}