Vfk thesis title. For details, please refer to the 2015 Proceedings of the National Training Team Lu Kaifeng, "The Properties and Application of Set Power Series and Its Fast Algorithm"
This problem is to find the set and convolution, you can use the fast Mobius transform.
The answer is ∑k=1ook∗(pk[U]−pk−1[U]) $\underset{k=1}{\overset{oo}{\sum }}k\ast (p^k[U]-p^{k-1}[U])$
After the Mobius transform, simplify it and you get a sequence of proportional numbers, because ∑p=1 $\underset{p}{\sum }=1$, So all but the last item are 0 −11−p $-\frac{1}{1-p}$
Then quickly Mobius inversion can return. The answer is U's coefficient.
complexity O(n2n) $O(n{2}^n)$

#include <bits/stdc++.h>
using namespace std;
#define ll long long
#define inf 0x3f3f3f3f
#define N 25
#define eps 1e-7
inline int read(){
    int x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-') f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9') x=x*10+ch-'0',ch=getchar();
    return x*f;
}
int n,bin[N];
double a[1100000];
int main(){
//  freopen("a.in","r",stdin);
    n=read();bin[0]=1;
    for(int i=1;i<=n;++i) bin[i]=bin[i-1]<<1;
    for(int i=0;i<bin[n];++i) scanf("%lf",&a[i]);
    for(int i=0;i<n;++i)
        for(int s=0;s<bin[n];++s)
            if(s&bin[i]) a[s]+=a[s^bin[i]];
    for(int i=0;i<bin[n];++i){
        if(a[i]>1-eps){
            if(i!=bin[n]-1){puts("INF");return 0;}
            a[i]=0;continue;
        }a[i]=1.0/(a[i]-1);
    }for(int i=0;i<n;++i)
        for(int s=0;s<bin[n];++s)
            if(s&bin[i]) a[s]-=a[s^bin[i]];
    printf("%.7lf\n",a[bin[n]-1]);
    return 0;
}