https://www.cnblogs.com/AwD-/p/6600650.html

#include<iostream> 
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<algorithm>
using namespace std;
#define ll long long
#define P 1000000007
#define N 1024
char getc(){char c=getchar();while ((c<‘A‘||c>‘Z‘)&&(c<‘a‘||c>‘z‘)&&(c<‘0‘||c>‘9‘)) c=getchar();return c;}
int gcd(int n,int m){return m==0?n:gcd(m,n%m);}
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;
}
int n,a[N],f[2][N][N];
void inc(int &x,int y){x+=y;if (x>=P) x-=P;}
int main()
{
#ifndef ONLINE_JUDGE
	freopen("a.in","r",stdin);
	freopen("a.out","w",stdout);
	const char LL[]="%I64d\n";
#else
	const char LL[]="%lld\n";
#endif
	n=read();
	for (int i=1;i<=n;i++) a[i]=read();
	f[0][1023][1023]=1;
	for (int i=0;i<n;i++)
	{
		for (int j=0;j<N;j++)
			for (int k=N-1^j;k>=0;k=k==0?-1:(k-1&(N-1^j)))
			f[i&1^1][j][k|j]=f[i&1][j][k|j];
		for (int j=0;j<N;j++)
			for (int k=N-1^j;k>=0;k=k==0?-1:(k-1&(N-1^j)))
			inc(f[i&1^1][j&a[i+1]][(k|j)&a[i+1]],f[i&1][j][k|j]),
			inc(f[i&1^1][j&a[i+1]][((k|j)&a[i+1])|j],P-f[i&1][j][k|j]);
	}
	cout<<f[n&1][0][0]<<endl;
	return 0;
}

BZOJ4762 最小集合（動態規劃+容斥原理）