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

题目大意：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

思路：
<摘自onepointo>
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
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];
可以发现当i==0时，dp[0]=B[0]/(1−A[0])；
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 <cstdio>#include <cstring>#include <algorithm>using namespace std;const int N = 510;double A[N], B[N];double p[100];int n, k1, k2, k3, a, b, c;void init(){    memset(p, 0, sizeof( p ));    memset(A, 0, sizeof( A ));    memset(B, 0, 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 = 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] += p0;//                }        for(int i=n; i>=0; --i){            A[i] = p0; B[i] = 1;            for(int j=1; j<=k1+k2+k3; ++j){                A[i] += A[i+j] * p[j];                B[i] += B[i+j] * p[j];            }        }        printf("%.16lf\n", B[0] / (1 - A[0]));    }    return 0;}