zoj3329 概率dp

One Person Game

---

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

---

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

/*
ZOJ 3329
题意：有三个骰子，分别有k1,k2,k3个面。
每次掷骰子，如果三个面分别为a,b,c则分数置0，否则加上三个骰子的分数之和。
当分数大于n时结束。求游戏的期望步数。初始分数为0

设dp[i]表示达到i分时到达目标状态的期望，pk为投掷k分的概率，p0为回到0的概率

//这里是逆向推导的过程，从i+k状态往i状态递推，有一次分值降为0则要多投一次

//概率dp真是博大精深，感觉还没入门
则dp[i]=∑(pk*dp[i+k])+dp[0]*p0+1;
都和dp[0]有关系，而且dp[0]就是我们所求，为常数
设dp[i]=A[i]*dp[0]+B[i];
代入上述方程右边得到：
dp[i]=∑(pk*A[i+k]*dp[0]+pk*B[i+k])+dp[0]*p0+1
     =(∑(pk*A[i+k])+p0)dp[0]+∑(pk*B[i+k])+1;
     明显A[i]=(∑(pk*A[i+k])+p0)
     B[i]=∑(pk*B[i+k])+1
     先递推求得A[0]和B[0].
     那么  dp[0]=B[0]/(1-A[0]);
*/

#include <iostream>
#include<stdio.h>
#include<string.h>
using namespace std;
double  A[600],B[600],p[600];
int main()
{
    double p0;
    int i,j,k,k1,k2,k3,a,b,c,m,n;
    scanf("%d",&n);
    while(n--)
    {
        scanf("%d%d%d%d%d%d%d",&m,&k1,&k2,&k3,&a,&b,&c);
        p0=1.0/k1/k2/k3;
        memset(p,0,sizeof(p));
        memset(A,0,sizeof(A));
        memset(B,0,sizeof(B));
        for(i=1;i<=k1;i++)
        for(j=1;j<=k2;j++)
        for(k=1;k<=k3;k++)
        if(i!=a||j!=b||k!=c)
        p[i+j+k]+=p0;
        for(i=m;i>=0;i--)
        {
            A[i]=p0;B[i]=1;
            for(j=1;j<=k1+k2+k3;j++)
            {
                A[i]+=A[j+i]*p[j];
                B[i]+=B[j+i]*p[j];
            }
        }
        printf("%.16f\n",B[0]/(1-A[0]));
    }
    return 0;
}