Tribles(概率)

Description

 

Problem A
Tribbles
Input: Standard Input

Output: Standard Output

GRAVITATION, n. "The tendency of all bodies to approach one another with a strength proportion to the quantity of matter they contain -- the quantity of matter they contain being ascertained by the strength of their tendency to approach one another. This is a lovely and edifying illustration of how science, having made A the proof of B, makes B the proof of A."

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.

Sample Input	Sample Output
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);

 1 #include<stdio.h>

 2 #include<math.h>

 3 #include<string.h>

 4 int main()

 5 {

 6     int test;

 7     scanf("%d",&test);

 8     for(int item = 1; item <= test; item++)

 9     {

10         int n,k,m;

11         double p[1100],f[1100];

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

13         for(int i = 0; i < n; i++)

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

15 

16         f[0] = 0;

17         f[1] = p[0];

18         for(int i = 2; i <= m; i++)

19         {

20             f[i] = 0;

21             for(int j = 0; j < n; j++)

22                 f[i] += p[j] * pow(f[i-1],j);

23         }

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

25     }

26     return 0;

27 }

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