bzoj 4036 [HAOI2015]按位或——min-max容斥+FMT

题目：https://www.lydsy.com/JudgeOnline/problem.php?id=4036

题解：https://www.cnblogs.com/Zinn/p/10260126.html

#include<cstdio>
#include<cstring>
#include<algorithm>
#define db double
using namespace std;
const int N=,M=(<<)+;
int n,bin[N],lm,ct[M];
db ans,mn[M],p[M];bool vis[N];
void fmt(db *a)
{
  for(int i=;i<lm;i<<=)
    for(int s=;s<lm;s++)
      if(s&i)p[s]+=p[s^i];
}
int main()
{
  scanf("%d",&n);
  bin[]=;for(int i=;i<=n;i++)bin[i]=bin[i-]<<; lm=bin[n];
  for(int s=;s<lm;s++)
    {
      scanf("%lf",&p[s]); if(!p[s])continue;
      for(int i=;i<n;i++)if(s&bin[i])vis[i]=;
    }
  for(int i=;i<n;i++)if(!vis[i]){puts("INF");return ;}
  fmt(p);
  for(int s=,u=lm-;s<lm;s++)mn[s]=/(-p[u^s]);//s=1//only p[u]=1 for legal
  for(int s=;s<lm;s++)ct[s]=ct[s-(s&-s)]+;
  for(int s=;s<lm;s++)//s=1
    (ct[s]&)?ans+=mn[s]:ans-=mn[s];
  printf("%.10f\n",ans);
  return ;
}