ZOJ 3329 One Person Game （期望DP）

One Person Game

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 3.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

題目大意：有三個骰子，分别有k1,k2,k3個面。

每次擲骰子，如果三個面分别為a,b,c則分數置0，否則加上三個骰子的分數之和。

當分數大于n時結束。求遊戲的期望步數。初始分數為0

思路：

<摘自onepointo>

設dp[i]表示達到i分時到達目标狀态的期望，pk為投擲k分的機率，p0為回到0的機率

則 dp[i]=∑(pk∗dp[i+k])+dp[0]∗p0+1

都和dp[0]有關系，而且dp[0]就是我們所求，為常數

設 dp[i]=A[i]∗dp[0]+B[i];

可以發現當i==0時，dp[0]=B[0]/(1−A[0])；

代入上述方程右邊得到：

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

明顯 A[i]=(∑(pk∗A[i+k])+p0)

B[i]=∑(pk∗B[i+k])+1

先遞推求得A[0]和B[0].

那麼 dp[0]=B[0]/(1−A[0]);

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;

const int N = ;

double A[N], B[N];
double p[];
int n, k1, k2, k3, a, b, c;

void init(){
    memset(p, , sizeof( p ));
    memset(A, , sizeof( A ));
    memset(B, , sizeof( B ));
}

int main(){
    int T; scanf("%d", &T);
    while( T-- ){
        init();
        scanf("%d%d%d%d%d%d%d", &n, &k1, &k2, &k3, &a, &b, &c);
        double p0 =  / k1 / k2 / k3;
        for(int i=; i<=k1; ++i)
            for(int j=; j<=k2; ++j)
                for(int k=; k<=k3; ++k){
                    if(i != a || j != b || k != c) p[i+j+k] += p0;//
                }
        for(int i=n; i>=; --i){
            A[i] = p0; B[i] = ;
            for(int j=; j<=k1+k2+k3; ++j){
                A[i] += A[i+j] * p[j];
                B[i] += B[i+j] * p[j];
            }
        }
        printf("%.16lf\n", B[] / ( - A[]));
    }
    return ;
}