传送门

题意：

思路：

直接求$i\&j>=k$不是很好求，所以转换成$i\&j=k$的情况。
考虑对$a,b$求一遍超集，让后从$[0,n-1]$扫一遍即可得到$i\&j=k$的情况，之后取一个后缀最大值，让后求和即可。

// Problem: K - I love max and multiply
// Contest: Virtual Judge - 2021多校第二场补题
// URL: https://vjudge.net/contest/448870#problem/K
// Memory Limit: 65 MB
// Time Limit: 1000 ms
// 
// Powered by CP Editor (https://cpeditor.org)//#pragma GCC optimize("Ofast,no-stack-protector,unroll-loops,fast-math")
//#pragma GCC target("sse,sse2,sse3,ssse3,sse4.1,sse4.2,avx,avx2,popcnt,tune=native")
//#pragma GCC optimize(2)
#include<cstdio>
#include<iostream>
#include<string>
#include<cstring>
#include<map>
#include<cmath>
#include<cctype>
#include<vector>
#include<set>
#include<queue>
#include<algorithm>
#include<sstream>
#include<ctime>
#include<cstdlib>
#include<random>
#include<cassert>
#define X first
#define Y second
#define L (u<<1)
#define R (u<<1|1)
#define pb push_back
#define mk make_pair
#define Mid ((tr[u].l+tr[u].r)>>1)
#define Len(u) (tr[u].r-tr[u].l+1)
#define random(a,b) ((a)+rand()%((b)-(a)+1))
#define db puts("---")
using namespace std;//void rd_cre() { freopen("d://dp//data.txt","w",stdout); srand(time(NULL)); }
//void rd_ac() { freopen("d://dp//data.txt","r",stdin); freopen("d://dp//AC.txt","w",stdout); }
//void rd_wa() { freopen("d://dp//data.txt","r",stdin); freopen("d://dp//WA.txt","w",stdout); }typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int,int> PII;const int N=1000010,mod=998244353,INF=0x3f3f3f3f;
const double eps=1e-6;int n;
int a[N],b[N];
LL mx1[N],mx2[N],mi1[N],mi2[N],c[N];int main()
{
//	ios::sync_with_stdio(false);
//	cin.tie(0);int _; scanf("%d",&_);while(_--) {scanf("%d",&n);for(int i=0;i<n;i++) scanf("%d",&a[i]);for(int i=0;i<n;i++) scanf("%d",&b[i]);int all=(int)log2(n)+1;for(int i=0;i<1<<all;i++) {mx1[i]=mx2[i]=-INF;mi1[i]=mi2[i]=INF;c[i]=-1e18;}for(int i=0;i<n;i++) mx1[i]=mi1[i]=a[i];for(int i=0;i<n;i++) mx2[i]=mi2[i]=b[i];for(int i=0;i<all;i++) {for(int j=0;j<1<<all;j++) {if(j>>i&1) {mx1[j^(1<<i)]=max(mx1[j^(1<<i)],mx1[j]);}}}for(int i=0;i<all;i++) {for(int j=0;j<1<<all;j++) {if(j>>i&1) {mi1[j^(1<<i)]=min(mx1[j^(1<<i)],mx1[j]);}}}for(int i=0;i<all;i++) {for(int j=0;j<1<<all;j++) {if(j>>i&1) {mx2[j^(1<<i)]=max(mx2[j^(1<<i)],mx2[j]);}}}for(int i=0;i<all;i++) {for(int j=0;j<1<<all;j++) {if(j>>i&1) {mi2[j^(1<<i)]=min(mi2[j^(1<<i)],mi2[j]);}}}LL now=1ll*INF*INF;for(int i=0;i<n;i++) {if(mx1[i]!=INF&&mx2[i]!=INF) c[i]=max(c[i],mx1[i]*mx2[i]);if(mi1[i]!=-INF&&mi2[i]!=-INF) c[i]=max(c[i],mi1[i]*mi2[i]); if(mx1[i]!=INF&&mi2[i]!=-INF) c[i]=max(c[i],mx1[i]*mi2[i]);if(mi1[i]!=-INF&&mx2[i]!=INF) c[i]=max(c[i],mi1[i]*mx2[i]);}for(int i=n-2;i>=0;i--) c[i]=max(c[i],c[i+1]);LL ans=0;for(int i=0;i<n;i++) ans+=c[i]%mod,ans+=mod,ans%=mod;printf("%lld\n",ans);}return 0;
}
/**/