Problem A
Tribbles
Input: Standard Input

Output: Standard Output

Ambrose Bierce

You have a population of kTribbles. This particular species of Tribbles live for exactly one day and then die. Just before death, a single Tribble has the probability P_i of giving birth to i more Tribbles. What is the probability that after m generations, everyTribble will be dead?

Input
The first line of input gives the number of cases, N. N test cases follow. Each one starts with a line containing n (1<= n<=1000) ,k (0<= k<=1000) and m (0<= m<=1000) . The next n lines will give the probabilities P₀, P₁, ...,P_n-1.

Output
For each test case, output one line containing "Case #x:" followed by the answer, correct up to an absolute or relative error of 10^-6.

4 

3 1 1

0.33 

0.34 

0.33 

3 1 2 

0.33 

0.34 

0.33 

3 1 2 

0.5 

0.0 

0.5 

4 2 2

0.5 

0.0 

0.0 

0.5

Case #1: 0.3300000 

Case #2: 0.4781370 

Case #3: 0.6250000 

Case #4: 0.3164062 

题意：有K只麻球，每只只活一天，临死前会产仔，产i只小麻球的 概率为pi,问m天后所有麻球全部死亡的概率；

思路：因为每只麻球都是相互独立的，所以只需求刚开始只有一只麻球，m天后其后代全部死亡的概率f[m]，然后k只麻球最后全部死亡的概率就是 pow(f[m],k);

对于一只麻球，m天全死亡包含第一天、第二天、、、、、第m天死亡事件，因此一只麻球第i天死亡的概率f[i] = p0 + p1*f[i-1] + p2*f[i-2]^2+.......+ pn-1*f[i-1]^(n-1);

 #include<stdio.h>

 #include<math.h>

 #include<string.h>

 int main()

 {

     int test;

     scanf("%d",&test);

     for(int item = ; item <= test; item++)

     {

         int n,k,m;

         double p[],f[];

         scanf("%d %d %d",&n,&k,&m);

         for(int i = ; i < n; i++)

             scanf("%lf",&p[i]);

         f[] = ;

         f[] = p[];

         for(int i = ; i <= m; i++)

         {

             f[i] = ;

             for(int j = ; j < n; j++)

                 f[i] += p[j] * pow(f[i-],j);

         }

         printf("Case #%d: %.7lf\n",item,pow(f[m],k));

     }

     return ;

 }