题意

\(N\)个物品，每次得到第\(i\)个物品的概率为\(p_i\)，而且有可能什么也得不到，问期望多少次能收集到全部\(N\)个物品

Sol

最直观的做法是直接状压，设\(f[sta]\)表示已经获得了\(sta\)这个集合里的所有元素，距离全拿满的期望，推一推式子直接转移就好了

主程序代码：

int N;double a[MAXN], f[MAXN];signed main() {//  freopen("a.in", "r", stdin);    while(scanf("%d", &N) != EOF) {         memset(f, 0, sizeof(f)); double res = 1.0;        for(int i = 0; i < N; i++) scanf("%lf", &a[i]), res -= a[i];        int Lim = (1 << N) - 1;        for(int sta = Lim - 1; sta >= 0; sta--) {            double now = 1 - res, sum = 0;            for(int i = 0; i < N; i++)                 if(sta & (1 << i)) now -= a[i];                else sum += f[sta | (1 << i)] * a[i];            sum += 1.0;            f[sta] = sum / now;        }        printf("%.4lf\n", f[0]);    }    return 0;}

另一种MinMax容斥的做法：

设\(max(s)\)为\(s\)集合中的最大元素，\(min(T)\)为集合\(T\)中的最小元素

那么有\(E(max(s)) =\sum_{T \subseteq S} (-1)^{|T| + 1} E(min \{ T \})\)

这里的\(E(max(S))\)显然就是我们要求的答案

\(E(min \{ T\}) = \frac{1}{\sum_{i \in T} p_i}\)

直接dfs一波

#include
    
      #define Pair pair
     
      #define MP(x, y) make_pair(x, y)#define fi first#define se second//#define int long long #define LL long long #define Fin(x) {freopen(#x".in","r",stdin);}#define Fout(x) {freopen(#x".out","w",stdout);}using namespace std;const int MAXN = 2e6 + 10, mod = 1e9 + 7, INF = 1e9 + 10;const double eps = 1e-7;template 
      
        inline bool chmin(A &a, B b){if(a > b) {a = b; return 1;} return 0;}template 
       
         inline bool chmax(A &a, B b){if(a < b) {a = b; return 1;} return 0;}template 
        
          inline LL add(A x, B y) {if(x + y < 0) return x + y + mod; return x + y >= mod ? x + y - mod : x + y;}template 
         
           inline void add2(A &x, B y) {if(x + y < 0) x = x + y + mod; else x = (x + y >= mod ? x + y - mod : x + y);}template 
          
            inline LL mul(A x, B y) {return 1ll * x * y % mod;}template 
           
             inline void mul2(A &x, B y) {x = (1ll * x * y % mod + mod) % mod;}template 
            
              inline void debug(A a){cout << a << '\n';}template 
             
               inline LL sqr(A x){return 1ll * x * x;}inline int read() { char c = getchar(); int x = 0, f = 1; while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();} while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar(); return x * f;}int N;double a[MAXN], ans;void dfs(int x, double p, int opt) { if(x == N) { if(p > eps) ans += opt / p; return ; } dfs(x + 1, p, opt); dfs(x + 1, p + a[x], -opt);}signed main() {// freopen("a.in", "r", stdin); while(scanf("%d", &N) != EOF) { for(int i = 0; i < N; i++) scanf("%lf", &a[i]); ans = 0; dfs(0, 0, -1); printf("%.4lf\n", ans); } return 0;}