Play the Dice

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65535/32768 K (Java/Others)

Total Submission(s): 3648    Accepted Submission(s): 1181

Special Judge

Problem Description

There is a dice with n sides, which are numbered from 1,2,...,n and have the equal possibility to show up when one rolls a dice. Each side has an integer ai on it. Now here is a game that you can roll this dice once, if the i-th side is up, you will get ai yuan. What's more, some sids of this dice are colored with a special different color. If you turn this side up, you will get once more chance to roll the dice. When you roll the dice for the second time, you still have the opportunity to win money and rolling chance. Now you need to calculate the expectations of money that we get after playing the game once.

 

 

Input

Input consists of multiple cases. Each case includes two lines.

The first line is an integer n (2<=n<=200), following with n integers a

_i(0<=a

_i<200)

The second line is an integer m (0<=m<=n), following with m integers b

_i(1<=b

_i<=n), which are the numbers of the special sides to get another more chance.

 

 

Output

Just a real number which is the expectations of the money one can get, rounded to exact two digits. If you can get unlimited money, print inf.

 

 

Sample Input

6 1 2 3 4 5 6 0 4 0 0 0 0 1 3

 

 

Sample Output

3.50 0.00

 

 

Source

 

 

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数学期望公式：E(X)=Xi乘Pi (i=1，2，3.....) X有几个值

 

给你一个有n个面的筛子，每个面都有一个值，另外有m面如果选中了，可以在投一次，问你期望是多少。

 

如果没有“m面选中可以在投一次”这一条件，那么期望就是n个面的值之和除以n，记为p=sum/n。

 

如果有了这个条件，那么只投一次的话，期望就是sum；如果投了两次的话就是p*(m/n);如果投了三次的话就是p*(m/n)^2,无限次就是p*(1+q+q^2+q^3+..+q^k+...）（q=m/n,p=sum/n)

 

根据等比数列以及无限化简得：sum/(n-m)。

 

 

1 #include 
     
       2 #include 
      
        3 #include 
       
         4 #include 
        
          5 using namespace std; 6 int main() 7 { 8     int n; 9     while (cin >> n)10     {11         int i;12         int s = 0;13         int x;14         for (i = 1; i <= n; i++)15         {16             cin >> x;17             s += x;18         }19         int m;20         cin >> m;21         for (i = 1; i <= m; i++)22         {23             cin >> x;24         }25         if (s == 0) cout << "0.00" << endl;//这不能省，因为n==m，s=0时，不是inf26         else if (n == m)27             cout << "inf" << endl;28         else printf("%.2lf\n", (double)s/ (n - m));29     }30     return 0;31 }