zoj3329（概率dp）

Description

There is a very simple and interesting one-person game. You have 3 dice, namely Die1, Die2 and Die3. Die1 has K₁ faces. Die2 hasK₂ 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 / K₂and 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：
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[n],b[n]; a[n-1],b[n-1] ... a[0],b[0];
则 E[0] = b[0]/(1 - a[0]);

#include <iostream>
#include <stdio.h>
#include <stdlib.h>
#include<string.h>
#include<algorithm>
#include<math.h>
#include<queue>
#include<iomanip>
using namespace std;
typedef long long ll;
int main()
{
    double p[20];
    int t;
    scanf("%d",&t);
    while(t--)
    {
        memset(p,0,sizeof(p));
        int n,k1,k2,k3,a,b,c;
        scanf("%d%d%d%d%d%d%d",&n,&k1,&k2,&k3,&a,&b,&c);
        double p0=1.0/(double)(k1*k2*k3);
        for(int i=1; i<=k1; i++)
            for(int j=1; j<=k2; j++)
                for(int k=1; k<=k3; k++)
                    if(i!=a||j!=b||k!=c)
                        p[i+j+k]+=p0;
        double A[520] = {0}, B[520] = {0};
        for(int i=n; i>=0; i--)
        {
            for(int k=3; k<=k1+k2+k3; k++)
            {
                A[i] +=A[i+k]*p[k];
                B[i] +=B[i+k]*p[k];
            }
            A[i]+=p0;
            B[i]++;
        }
        printf("%.15lf\n", B[0]/(1 - A[0]) );
    }
    return 0;
}