---

考虑min-max容斥

\(E[max(S)] = \sum \limits_{T \subset S} min(T)\)

\(min(T)\)是可以被表示出来

即所有与\(T\)有交集的数的概率的和的倒数

通过转化一下，可以考虑求所有与\(T\)没有交集的数的概率和

即求\(T\)的补集的子集的概率和

用FMT随意做下吧...

注意：概率为1的时候需要特判

复杂度\(O(2^n * n)\)

---

#include <cstdio>

#include <vector>

#include <cstring>

#include <iostream>

#include <algorithm>

using namespace std;

#define de double

#define ri register int

#define rep(io, st, ed) for(ri io = st; io <= ed; io ++)

#define drep(io, ed, st) for(ri io = ed; io >= st; io --)

const int sid = (1 << 20) + 25;

int n, show;

de Max, sub[sid];

int main() {

    scanf("%d", &n);

    rep(i, 0, (1 << n) - 1) {

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

        show |= i * (sub[i] > 1e-8);

    }

    if(show != (1 << n) - 1) { puts("INF"); return 0; }

    rep(i, 1, n) rep(S, 0, (1 << n) - 1)

        if(!(S & (1 << i - 1)))

            sub[S ^ (1 << i - 1)] += sub[S];

    int T = (1 << n) - 1;

    rep(S, 1, (1 << n) - 1) { // no 0

        if(__builtin_popcount(S) & 1) Max += 1.0 / (1.0 - sub[T ^ S]);

        else Max -= 1.0 / (1.0 - sub[T ^ S]);

    }

    printf("%.12lf\n", Max);

    return 0;

}