BZOJ4036 HAOI2015按位或（概率期望+容斥原理）

　　考虑min-max容斥，改为求位集合内第一次有位变成1的期望时间。求出一次操作选择了S中的任意1的概率P[S]，期望时间即为1/P[S]。

　　考虑怎么求P[S]。P[S]=∑p[s] (s&S>0)=1-∑p[s] (s&S==0)。做一个高维前缀和即可。

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<cstdlib>
#include<algorithm>
using namespace std;
#define N (1<<20)
int read()
{
    int x=0,f=1;char c=getchar();
    while (c<'0'||c>'9') {if (c=='-') f=-1;c=getchar();}
    while (c>='0'&&c<='9') x=(x<<1)+(x<<3)+(c^48),c=getchar();
    return x*f;
}
const double eps=1E-8;
int n,t,s[N];
double p[N],ans;
int main()
{
#ifndef ONLINE_JUDGE
    freopen("bzoj4036.in","r",stdin);
    freopen("bzoj4036.out","w",stdout);
    const char LL[]="%I64d
";
#else
    const char LL[]="%lld
";
#endif
    n=read();
    for (int i=0;i<(1<<n);i++)
    {
        scanf("%lf",&p[i]);
        if (p[i]>eps) t|=i;
    }
    if (t!=(1<<n)-1) {cout<<"INF";return 0;}
    for (int i=0;i<n;i++)
        for (int j=0;j<(1<<n);j++)
        if (j&(1<<i)) p[j]+=p[j^(1<<i)];
    for (int i=1;i<(1<<n);i++)
    {
        s[i]=s[i^(i&-i)]+1;
        if (s[i]&1) ans+=1/(1-p[(1<<n)-1^i]);
        else ans-=1/(1-p[(1<<n)-1^i]);
    }
    printf("%.6f",ans);
}