题意：有 k 只小鸟，每只都只能活一天，但是每只都可以生出一些新的小鸟，生出 i 个小鸟的概率是 Pi，问你 m 天所有的小鸟都死亡的概率是多少。
析：先考虑只有一只小鸟，dp[i] 表示 i 天全部死亡的概率，那么 dpi] = P0 + P1*dp[i-1] + P2*dp[i-1]^2 + ... + Pn*dp[i-1]^(n-1)，式子 Pjdp[i-1]^j 表示该小鸟生了 j 后代，，它们在 i-1 天死亡的概率是 dp[i-1]，因为有 j 只，每只都是 dp[i-1]，所以就是 dp[i-1]^j。
代码如下：
#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#include <sstream>
#include <list>
#include <assert.h>
#include <bitset>
#include <numeric>
#define debug() puts("++++")
#define gcd(a, b) __gcd(a, b)
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define fi first
#define se second
#define pb push_back
#define sqr(x) ((x)*(x))
#define ms(a,b) memset(a, b, sizeof a)
#define sz size()
#define pu push_up
#define pd push_down
#define cl clear()
#define lowbit(x) -x&x
//#define all 1,n,1
#define FOR(i,x,n) for(int i = (x); i < (n); ++i)
#define freopenr freopen("in.in", "r", stdin)
#define freopenw freopen("out.out", "w", stdout)
using namespace std; typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const LL LNF = 1e17;
const double inf = 1e20;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 1000 + 10;
const int maxm = 100 + 2;
const LL mod = 100000000;
const int dr[] = {-1, 1, 0, 0, 1, 1, -1, -1};
const int dc[] = {0, 0, 1, -1, 1, -1, 1, -1};
const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline bool is_in(int r, int c) {
return r >= 0 && r < n && c >= 0 && c < m;
} double p[maxn], dp[maxn]; int main(){
int T, k; cin >> T;
for(int kase = 1; kase <= T; ++kase){
scanf("%d %d %d", &n, &k, &m);
for(int i = 0; i < n; ++i) scanf("%lf", p + i);
dp[0] = 0; dp[1] = p[0];
for(int i = 2; i <= m; ++i){
dp[i] = p[0];
for(int j = 1; j < n; ++j) dp[i] += p[j] * pow(dp[i-1], j);
}
printf("Case #%d: %.6f\n", kase, pow(dp[m], k));
}
return 0;
}