【Codeforces 449D】Jzzhu and Numbers（高维前缀和）

传送门

先用高维前缀和求出$i$的超集的个数$a_i$
然后$2^{a_i}-1$就是$&$之和为$i$的超集的个数
然后高维前缀差分就得到为$i$的答案

#include<bits/stdc++.h>
using namespace std;
#define cs const
#define pb push_back
#define re register
#define ll long long
#define pii pair<int,int>
#define fi first
#define se second
cs int RLEN=1<<20|1;
inline char gc(){
	static char ibuf[RLEN],*ib,*ob;
	(ob==ib)&&(ob=(ib=ibuf)+fread(ibuf,1,RLEN,stdin));
	return (ib==ob)?EOF:*ib++;
}
inline int read(){
	char ch=gc();
	int res=0;bool f=1;
	while(!isdigit(ch))f^=ch=='-',ch=gc();
	while(isdigit(ch))res=(res+(res<<2)<<1)+(ch^48),ch=gc();
	return f?res:-res;
}
cs int mod=1e9+7;
template<class tp>inline void chemx(tp &a,tp b){a<b?a=b:0;}
template<class tp>inline void chemn(tp &a,tp b){a>b?a=b:0;}
inline void Add(int &a,int b,int mod){(a+=b)>=mod?(a-=mod):0;}
inline void Dec(int &a,int b,int mod){a-=b,a+=a>>31&mod;}
cs int N=(1<<20)|1;
int n,a[N],pw[N];
inline void fmt(int *f,int lim,int kd){
	for(int mid=1;mid<lim;mid<<=1)
	for(int i=0;i<lim;i+=mid<<1)
	for(int j=0;j<mid;j++)
	kd?Add(f[i+j],f[i+j+mid],mod-1):Dec(f[i+j],f[i+j+mid],mod);
}
int main(){
	#ifdef Stargazer
	freopen("lx.cpp","r",stdin);
	#endif
	n=read();pw[0]=1;
	for(int i=1;i<N-1;i++)pw[i]=(pw[i-1]+pw[i-1])%mod;
	for(int i=1;i<=n;i++)a[read()]++;
	fmt(a,N-1,1);
	for(int i=0;i<N-1;i++)a[i]=pw[a[i]];
	fmt(a,N-1,0);
	cout<<a[0];
}