ZOJ-3329 One Person Game（期望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

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

题解：设p0=1/(k1*k2*k3)为置零的概率，pk为增加k分的概率
设dp[i]总分为i时的期望值，则当i>n时，dp[i]=0

递推公式：dp[i]=∑pk*dp[i+k]+p0*dp[0]+1

由于展开后dp[i]只与dp[0]有关，且dp[0]是我们所要求的，为常数
因此可以设dp[i]=A[i]*dp[0]+B[i]

代入递推公式：dp[i]=∑pk*(A[i+k]*dp[0]+B[i+k])+p0*dp[0]+1
                          =(p0+∑pk*A[i+k])dp[0]+(1+∑pk*B[i+k])
                    
得到A[i]=p0+∑pk*A[i+k],B[i]=1+∑pk*B[i+k]

又由于dp[0]=A[0]*dp[0]+B[0],只要知道A[0]和B[0]就可以得到dp[0]

#include<bits/stdc++.h>
using namespace std;
double p[20],A[555],B[555];
int n,k1,k2,k3,a,b,c;
int main(){
    int T;
    //freopen("in.txt","r",stdin);
    scanf("%d",&T);
    while(T--){
        scanf("%d%d%d%d%d%d%d",&n,&k1,&k2,&k3,&a,&b,&c);
        int tot=k1+k2+k3;
        memset(p,0,sizeof(p));
        p[0]=1.0/(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]+=p[0];
        memset(A,0,sizeof(A));
        memset(B,0,sizeof(B));
        for(int i=n;i>=0;i--){
            A[i]=p[0];
            B[i]=1;
            for(int j=1;j<=tot;j++){
                A[i]+=p[j]*A[i+j];
                B[i]+=p[j]*B[i+j];
            }
        }
        printf("%.8f\n",B[0]/(1-A[0]));
    }
    return 0;
}