一道不知道哪里来的容斥题

题意:

容斥二进制意义下至少哪些位置不相等。

#include<iostream>
#include<cctype>
#include<cstdio>
#include<cstring>
#include<string>
#include<cmath>
#include<ctime>
#include<cstdlib>
#include<algorithm>
#define N 2200000
#define L 2000000
#define eps 1e-7
#define inf 1e9+7
#define db double
#define ll long long
#define ldb long double
using namespace std;
inline ll read()
{
	char ch=0;
	ll x=0,flag=1;
	while(!isdigit(ch)){ch=getchar();if(ch=='-')flag=-1;}
	while(isdigit(ch)){x=(x<<3)+(x<<1)+ch-'0';ch=getchar();}
	return x*flag;
}
const ll mo=998244353;
ll ksm(ll x,ll k)
{
	ll ans=1;
	while(k)
	{
		if(k&1)ans=ans*x%mo;
		k>>=1;
		x=x*x%mo;
	}
	return ans;
}
ll a[N],f[N],mi[N];
int main()
{
	ll n=read(),ans;
	mi[0]=1;
	for(ll i=1;i<=n;i++)a[i]=read(),mi[i]=(mi[i-1]*2)%mo;
	for(ll i=1;i<=(1<<20)-1;i++)f[i]=f[i>>1]+(i&1);
	ans=mi[n]-2;ans=(ans*ksm(2,mo-2))%mo;
	for(ll s=0;s<=(1<<20)-1;s++)
	{
		ll cnt=0;
		for(ll i=1;i<=n;i++)if(s==(a[i]&s))cnt++;
		if(cnt==0||cnt==n)continue;
		if(f[s]&1)ans=(ans-(mi[cnt]-1))%mo;
		else ans=(ans+(mi[cnt]-1))%mo;
	}
	printf("%lld",(ans%mo+mo)%mo);
	return 0;
}