NOIP 模拟 $36\; \rm Cicada 拿衣服$

#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();}