ZOJ-3329 One Person Game （概率DP）

One Person Game

http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=3754

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

题目大意：有三个分别有k1,k2,k3面的骰子，计数器初始为0，每次加上三个骰子点数和，若k1==a&&k2==b&&k3==c，则计数器清零，求最终点数超过n的掷骰子次数的期望？

设dp[i]表示点数i离目标状态还需掷骰子次数的期望

状态转移方程很好想：dp[i]=∑(p[k]*dp[i+k])+1/(k1*k2*k3)*dp[0]+1;                                                                ①

但是由于存在dp[0]，所以没法直接使用

刚开始我用dp[i]-dp[i+1]求出一个方程，然而右边常数项没了，就没法进行转移。。。

看了题解后才发现：由于dp[i]与dp[0]都有线性关系，所以可以设系数a[i],b[i]，令dp[i]=a[i]*dp[0]+b[i];      ②

将②式带入①式右边可得：dp[i]=∑(p[k]*(a[i+k]*dp[0]+b[i+k]))+1/(k1*k2*k3)*dp[0]+1;

整理可得：dp[i]=∑(p[k]*a[i+k]+1/(k1*k2*k3))*dp[0]+∑(p[k]*b[i+k])+1;                                                             ③

对比②③右边可得：

a[i]=∑(p[k]*a[i+k]+1/(k1*k2*k3));

b[i]=∑(p[k]*b[i+k])+1;

【上述各式中：3<=k&&k<=k1*k2*k3】

而dp[0]=a[0]*dp[0]+b[0] 可以推出：dp[0]=b[0]/(1-a[0]);

所以可以先推出a[0]和b[0]，再求出dp[0];

感觉概率dp就是数列求能够直接递推的递推公式

#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;

int n,k1,k2,k3,x,y,z;
double a[625],b[625],p[225],sum;

int main() {
    int T;
    scanf("%d",&T);
    while(T-->0) {
        scanf("%d%d%d%d%d%d%d",&n,&k1,&k2,&k3,&x,&y,&z);
        memset(p,0,sizeof(p));
        memset(a,0,sizeof(a));
        memset(b,0,sizeof(b));
        for(int i=1;i<=k1;++i) {
            for(int j=1;j<=k2;++j) {
                for(int k=1;k<=k3;++k) {
                    p[i+j+k]+=1;
                }
            }
        }
        sum=k1*k2*k3;
        p[0]=1;
        p[x+y+z]-=1;
        for(int i=sum;i>=0;--i) {
            p[i]/=sum;
        }
        for(int i=n;i>=0;--i) {
            a[i]=p[0];
            b[i]=1;
            for(int k=3;k<=sum;++k) {
                a[i]+=p[k]*a[i+k];
                b[i]+=p[k]*b[i+k];
            }
        }
        printf("%.15lf\n",b[0]/(1-a[0]));
    }
    return 0;
}