[Mo] + bitset team Luo Gu P4688

Title whitewash prime sunny day In fact, because parents at home can not play Su sunny day

We ran out of the most inferior solutions often do not want the card, write a solution to a problem in memory of ......

Of course, it is the first discrete look, if only then consider seeking the kind of the same color range, then how do

You can open \ (\ texttt {bitset} \ ) recording each color not occurred, then three sections \ (\ texttt {bitset} \ ) were combined, the combined answer is \ (\ texttt {bitset} \ ) in \ (1 \) number

Each of the processing sections \ (\ texttt {bitset} \ ) can team Mo

Merge found the answer space will be fried, so consider asking grouping, I probably every \ (5000 \) one asks in a group run

Good surface back to the original title, the same three intervals required seeking the number of color number. Not as long as the weight of discrete, with \ (nxt [x] \) Recording (X \) \ next in \ (\ texttt {bitset} \ ) in a position on the line, or above that practice.

#include <bits/stdc++.h>
#define For(i,x,y) for(int i=(x);i<=(y);++i)
#define FoR(i,x,y,z) for(int i=(x);i<=(y);i+=(z))
#define Rof(i,x,y) for(int i=(x);i>=(y);--i)
#define RoF(i,x,y,z) for(int i=(x);i>=(y);i-=(z))
#define Edge(x) for(int i=head[x],to=e[i].v;i;i=e[i].nxt,to=e[i].v)
typedef long long ll;
const int N=100005;
const int M=5000;
using namespace std;

template<typename T> void rd(T& x){
    T y=0;int f=1;char c=getchar();
    while(c<'0' || c>'9')
    { c=getchar(); if(c=='-')f=-f; }
    while(c>='0' && c<='9') y=y*10+c-'0',c=getchar();
    x=y*f;
}

int a[N],b[N],bl[N],cnt,nxt[N],tim[N],ans[N],L[N][3],R[N][3],n;
bitset<N> tmp,p[M+2];
struct qwq{
    int l,r,id;
    bool operator <(const qwq &x)const
    { return bl[l]==bl[x.l]?(bl[l]&1?r<x.r:r>x.r):bl[l]<bl[x.l]; }
}q[3*M+5];

void add(int t,int x)
{ x=a[x]; if(tim[x]<t) tim[x]=t,nxt[x]=x; tmp[nxt[x]++]=1; }
void del(int x)
{ x=a[x],tmp[--nxt[x]]=0; }
void solve(int t){
    int l=1,r=0; tmp.reset();
    sort(q+1,q+cnt+1);
    For(i,1,cnt){
        while(l>q[i].l) add(t,--l);
        while(r<q[i].r) add(t,++r);
        while(l<q[i].l) del(l++);
        while(r>q[i].r) del(r--);
        p[q[i].id]&=tmp;
    }
    For(i,0,M-1){
        if(t*M+i>n) return;
        ans[t*M+i]-=3*p[i].count();
    }
}

int main(){
//  freopen("in.txt","r",stdin);
//  freopen("wa.txt","w",stdout);
    int m; rd(n),rd(m);
    int block=sqrt(n);
    For(i,1,n) rd(a[i]),b[i]=a[i]; sort(b+1,b+n+1);
    For(i,1,n) a[i]=lower_bound(b+1,b+n+1,a[i])-b;
    For(i,1,n) bl[i]=(i-1)/block+1;
    
    For(i,0,m-1) For(j,0,2){
        rd(L[i][j]),rd(R[i][j]);
        ans[i]+=R[i][j]-L[i][j]+1;
    }
    memset(tim,-1,sizeof(tim));
    For(i,0,(m+M-1)/M){
        cnt=0;
        For(j,i*M,min(m-1,i*M+M-1)){
            p[j%M].set();
            For(k,0,2) q[++cnt]=(qwq){L[j][k],R[j][k],j%M};         
        }
        solve(i);
    }
    For(i,0,m-1) printf("%d\n",ans[i]);
}