zoj3329 One Person Game

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, namelyDie1, Die2 and Die3. Die1 hasK₁ faces. Die2 has K₂ faces.Die3 has K₃ faces. All the dice are fair dice, so the probability of rolling each value, 1 toK₁, 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 isa, the up-facing number of Die2 is b and the up-facing number ofDie3 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 integersn, 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

20 2 2 2 1 1 10 6 6 6 1 1 1

Sample Output

1.1428571428571431.004651162790698

我们可以推出公式dp[i]=sum{pk*dp[i+k]}+p0*dp[0]+1,dp[i]表示当前分数是i是时的到游戏结束的期望，那么dp[0],就是要求的，其实，我们可以发现一个归律，求期望都是从后住前推的，比如，这里，是由i+k推到i的，求概率的时候是刚好相反的，但我们发现求出的这个式子，是个环形的，所以要变形，因为，每一个都是和dp[0]相关的，我们可以设，dp[i]=a[i]*dp[0]+b[i],那么dp[0]不就是，b[i]/(1-a[i])了么，我们，把这个式子代入上式就可以得到，dp[i]=sum{pk*a[i+k]}+p0,b[i]=sum{pk*b[i+k]}+1,这样，就可以马上求结果来了！

#include <iostream>#include <stdio.h>#include <string.h>using namespace std;#define MAXN 550double pa[MAXN],pb[MAXN],dp[MAXN],p[MAXN];int main(){    int tcase,i,j,k,n,a,b,c,k1,k2,k3;    double p0;    scanf("%d",&tcase);    while(tcase--)    {        scanf("%d%d%d%d%d%d%d",&n,&k1,&k2,&k3,&a,&b,&c);        p0=1.0/k1/k2/k3;        memset(pa,0,sizeof(pa));        memset(pb,0,sizeof(pb));        memset(dp,0,sizeof(dp));        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)//不同时相等                    p[i+j+k]+=p0;        int temp=k1+k2+k3;        for(i=n;i>=0;i--)        {            pa[i]=p0;            pb[i]=1;            for(k=1;k<=temp;k++)            {                pa[i]+=pa[i+k]*p[k];                pb[i]+=pb[i+k]*p[k];            }        }        printf("%.15f\n",pb[0]/(1.0-pa[0]));    }    return 0;}