11021 - Tribles

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 k Tribbles. This particular species of Tribbles live for exactly one day and

then die. Just before death, a single Tribble has the probability Pi of giving birth to i more Tribbles.

What is the probability that after m generations, every Tribble 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, . . . , Pn−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

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

Sample Output

Case #1: 0.3300000

Case #2: 0.4781370

Case #3: 0.6250000

Case #4: 0.3164062

题解：求概率，由全概率公式可以得到f[i]=p[0]+p[1]f[i-1]+p[2]f[i-1]^2+......+p[n-1]f[i-1]^[n-1];

题意就是给你一系列概率代表生i个麻球的概率，让求m天后都死亡的概率；This particular species of Tribbles live for exactly one day and

then die.说明包括中途死的；p[j]*f[i-1]^j代表这个麻球生了j个后代，他们在I-1天全死亡的概率；由于是独立的所以是j次方；

最后结果还要k次方；

代码：

#include
     
      #include
      
       #include
       
        #include
        
         #include
         
          using namespace std;#define mem(x,y) memset(x,y,sizeof(x))const int MAXN=1010;double p[MAXN],f[MAXN];int main(){	int T,n,k,m,kase=0;	scanf("%d",&T);	while(T--){		scanf("%d%d%d",&n,&k,&m);		for(int i=0;i