zoj 3329 One Person Game (概率DP )

One Person Game

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Time Limit: 1 Second      Memory Limit: 32768 KB      Special Judge

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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

思路：dp[i]=∑(dp[i+k]*p[k])+dp[0]*p0+1;  (k=3..k1+k2+k3) //dp[i]：从该点走到游戏结束的期望步数。

由于对于固定的k1、k2、k3，求得dp[0]是一个定值。

能够设dp[i]=A[i]*dp[0]+B[i]; 带入上式能够得到

dp[i]=(∑(p[k]*A[i+k])+p0)dp[0]+∑(p[k]*A[i+k])+p0;

A[i]=∑(p[k]*A[i+k])+p0;

B[i]=∑(p[k]*B[i+k])+p0;

则：dp[0]=B[0]/(1-A[0]);

#include<stdio.h>
#include<math.h>
#include<string.h>
#include<stdlib.h>
#include<algorithm>
#include<iostream>
using namespace std;
#define N 550        //数组不能开小了!!
#define LL __int64
const int inf=0x1f1f1f1f;
const double eps=1e-10;
double p[20],A[N],B[N],p0;
int main()
{
    int T,i,j,k,n,a,b,c,k1,k2,k3;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d%d%d%d%d%d%d",&n,&k1,&k2,&k3,&a,&b,&c);
        double p0=1.0/(k1*k2*k3);
        memset(p,0,sizeof(p));
        for(i=1; i<=k1; i++)
        {
            for(j=1; j<=k2; j++)
            {
                for(k=1; k<=k3; k++)
                {
                    if(i==a&&j==b&&k==c)
                        continue;
                    p[i+j+k]+=p0;
                }
            }
        }
        memset(A,0,sizeof(A));
        memset(B,0,sizeof(B));
        for(i=n; i>=0; i--)
        {
            A[i]=p0;
            B[i]=1;
            for(j=1; j<=(k1+k2+k3); j++)
            {
                A[i]+=A[i+j]*p[j]; 
                B[i]+=B[i+j]*p[j];
            }
        }
        printf("%.10f
",B[0]/(1.0-A[0]));
    }
    return 0;
}