题解 \(by\;zj\varphi\)
发现右端点固定时,左端点的 \(min-max\) 单调递减,且对于 \(or\) 和 \(and\) 相减,最多有 \(\rm2logn\)个不同的值,且相同的值构成一段连续的区间。
那么就可以在最远的,符合答案的第一个区间二分答案。
具体实现可以用一个链表,每次扫一遍合并,并倒着查合法区间,这样就能做到 \(\rm nlogn\)。
Code
#include<bits/stdc++.h>
#define ri register signed
#define Re register
#define p(i) ++i
namespace IO{
char buf[1<<21],*p1=buf,*p2=buf;
#define gc() p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<21,stdin),p1==p2)?(-1):*p1++
struct nanfeng_stream{
template<typename T>inline nanfeng_stream operator>>(T &x) {
ri f=0;x=0;register char ch=gc();
while(!isdigit(ch)) f|=ch=='-',ch=gc();
while(isdigit(ch)) x=(x<<1)+(x<<3)+(ch^48),ch=gc();
return x=f?-x:x,*this;
}
}cin;
}
using IO::cin;
namespace nanfeng{
#define node(x,y,z) (node){x,y,z}
#define eb emplace_back
#define FI FILE *IN
#define FO FILE *OUT
template<typename T>inline T cmax(T x,T y) {return x>y?x:y;}
template<typename T>inline T cmin(T x,T y) {return x>y?y:x;}
static const int N=1e6+7;
struct node{int l,OR,AND;int val() {return OR-AND;}};
std::list<node> lis;
int a[N],stn[N][23],stx[N][23],lg[N],n,k;
inline void init_rmq() {
for (ri i(1);i<=n;p(i)) stx[i][0]=stn[i][0]=a[i];
for (ri i(2);i<=n;p(i)) lg[i]=lg[i>>1]+1;
for (ri j(1);j<19;p(j)) {
ri len=1<<j;
for (ri i(1);i+len-1<=n;p(i)) {
stx[i][j]=cmax(stx[i][j-1],stx[i+(1<<j-1)][j-1]);
stn[i][j]=cmin(stn[i][j-1],stn[i+(1<<j-1)][j-1]);
}
}
}
inline int Getmx(int l,int r) {
int k=lg[r-l+1];
return cmax(stx[l][k],stx[r-(1<<k)+1][k]);
}
inline int Getmn(int l,int r) {
int k=lg[r-l+1];
return cmin(stn[l][k],stn[r-(1<<k)+1][k]);
}
inline bool check(Re std::list<node>::iterator it,int l,int r) {return it->val()+Getmn(l,r)-Getmx(l,r)>=k;}
struct Seg{
#define ls(x) (x<<1)
#define rs(x) (x<<1|1)
struct segmenttree{int mx,lz;segmenttree(){mx=lz=-1;}}T[N<<2];
inline void down(int x) {
int l=ls(x),r=rs(x);
T[l].mx=cmax(T[l].mx,T[x].lz);
T[r].mx=cmax(T[r].mx,T[x].lz);
T[l].lz=cmax(T[l].lz,T[x].lz);
T[r].lz=cmax(T[r].lz,T[x].lz);
}
void update(int x,int l,int r,int lt,int rt) {
if (l<=lt&&rt<=r) {
T[x].lz=cmax(T[x].lz,r-l+1);
T[x].mx=cmax(T[x].mx,r-l+1);
return;
}
down(x);
int mid(lt+rt>>1);
if (l<=mid) update(ls(x),l,r,lt,mid);
if (r>mid) update(rs(x),l,r,mid+1,rt);
}
void print(int x,int l,int r) {
if (l==r) return (void)(printf("%d ",T[x].mx));
int mid(l+r>>1);
down(x);
print(ls(x),l,mid);
print(rs(x),mid+1,r);
}
}T;
inline int main() {
//FI=freopen("nanfeng.in","r",stdin);
//FO=freopen("nanfeng.out","w",stdout);
cin >> n >> k;
for (ri i(1);i<=n;p(i)) cin >> a[i];
init_rmq();
for (ri i(1);i<=n;p(i)) {
for (Re auto it=lis.begin();it!=lis.end();p(it)) it->OR|=a[i],it->AND&=a[i];
lis.eb(node(i,a[i],a[i]));
for (Re auto it1=lis.begin(),it2=next(it1);it2!=lis.end();p(it2))
if (it1->val()==it2->val()) lis.erase(it1),it1=it2;
else p(it1);
for (Re auto it=lis.begin();it!=lis.end();p(it))
if (check(it,it->l,i)) {
ri l=1,r=it->l,res;
if (it!=lis.begin()) l=prev(it)->l+1;
while(l<=r) {
int mid(l+r>>1);
if (check(it,mid,i)) r=mid-1,res=mid;
else l=mid+1;
}
T.update(1,res,i,1,n);
break;
}
}
T.print(1,1,n);
return 0;
}
}
int main() {return nanfeng::main();}