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<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<algorithm>
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<n;i++)scanf("%lf",p+i);
mem(f,0);
f[1]=p[0];f[0]=0;
for(int i=2;i<=m;i++){
for(int j=0;j<n;j++){
f[i]+=p[j]*pow(f[i-1],j);
}
}
printf("Case #%d: %.7lf\n",++kase,pow(f[m],k));
}
return 0;
}