zoj 3329 DP 求期望

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

double p[510];//dp[i]中dp[0]的系数
double con[510];//dp[i]的常数
//dp[i]=p[i]*dp[0]+con[i];
//dp[i]表示和为i时还需扔筛子的次数
//可以列出n个方程
/*
dp[n]=dp[0]*x+sigma(dp[n+k]*sum[k]);
dp[n-1]=dp[0]*x+sigma(dp[n+k]*sum[k]);
.....
dp[0]=dp[0]*x+sigma(dp[0+k]*sum[k]);
其中k为一次可能的筛子数的总和，sum[k]为和为k的概率

View Code

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
double sum[100];
double p[510];//dp[i]中dp[0]的系数
double con[510];//dp[i]的常数
int main()
{
     int t,n,k1,k2,k3,i,j,k,a,b,c;
     scanf("%d",&t);
     while(t--)
     {
          scanf("%d%d%d%d%d%d%d",&n,&k1,&k2,&k3,&a,&b,&c);
          double x=(1.0/k1*1.0/k2*1.0/k3);
          memset(p,0,sizeof(p));
          memset(con,0,sizeof(con));
          memset(sum,0,sizeof(sum));
          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;
                      sum[i+j+k]+=x;
                  }
              }
          }
          for(i=n;i>=0;i--)
          { 
              for(j=i+3;j<=k1+k2+k3+i;j++)
              {
                  p[i]+=p[j]*sum[j-i];
                  con[i]+=con[j]*sum[j-i];
              }
              p[i]+=x;
              con[i]+=1.0;
          }
          printf("%.15lf\n",con[0]/(1.0-p[0]));
     }
}