F. Machine Learning 带修端点莫队

F. Machine Learning

time limit per test

4 seconds

memory limit per test

512 megabytes

input

standard input

output

standard output

You come home and fell some unpleasant smell. Where is it coming from?

You are given an array a. You have to answer the following queries:

1. You are given two integers l and r. Let c_i be the number of occurrences of i in a_l: r, where a_l: r is the subarray of a from l-th element to r-th inclusive. Find the Mex of {c₀, c₁, ..., c_10⁹}
2. You are given two integers p to x. Change a_p to x.

The Mex of a multiset of numbers is the smallest non-negative integer not in the set.

Note that in this problem all elements of a are positive, which means that c₀ = 0 and 0 is never the answer for the query of the second type.

Input

The first line of input contains two integers n and q (1 ≤ n, q ≤ 100 000) — the length of the array and the number of queries respectively.

The second line of input contains n integers — a₁, a₂, ..., a_n (1 ≤ a_i ≤ 10⁹).

Each of the next q lines describes a single query.

The first type of query is described by three integers t_i = 1, l_i, r_i, where 1 ≤ l_i ≤ r_i ≤ n — the bounds of the subarray.

The second type of query is described by three integers t_i = 2, p_i, x_i, where 1 ≤ p_i ≤ n is the index of the element, which must be changed and 1 ≤ x_i ≤ 10⁹ is the new value.

Output

For each query of the first type output a single integer  — the Mex of {c₀, c₁, ..., c_10⁹}.

Example

Input

Copy

10 41 2 3 1 1 2 2 2 9 91 1 11 2 82 7 11 2 8

Output

#include<cstdio>
#include<cstring>
#include<map>
#include<cmath>
#include<algorithm>
using namespace std;
const int N=2e5+88;
map<int,int>M;
int vis[N],num[N],a[N],b[N],now[N],ans[N],unit,l,r,t;
struct Query{
    int l,r,tim,id;
    bool operator < (const Query &A)const{
      return l/unit==A.l/unit?(r/unit==A.r/unit?tim<A.tim:r<A.r):l<A.l;
    }
}Q[N];
struct Change{
    int pos,x,y;
}C[N];
void modify(int x,int d){
    --vis[num[x]];num[x]+=d;++vis[num[x]];
}
void going(int T,int d){
    if(C[T].pos>=l&&C[T].pos<=r) modify(C[T].x,d),modify(C[T].y,-d);
    if(d==1) a[C[T].pos]=C[T].x;else a[C[T].pos]=C[T].y;
}
int calc(){
    for(int i=1;;++i) if(!vis[i]) return i;
}
int main(){
    int n,q,tot,op,cc=0,pp=0;
    scanf("%d%d",&n,&q);
    for(int i=1;i<=n;++i) scanf("%d",&a[i]),now[i]=b[i]=a[i];
    tot=n,unit=(int)pow(n,0.6666666);
    for(int i=1;i<=q;++i) {
        scanf("%d",&op);
        if(op==2){
        ++cc,scanf("%d%d",&C[cc].pos,&C[cc].x);
        C[cc].y=now[C[cc].pos],b[++tot]=now[C[cc].pos]=C[cc].x;
        }
        else {
        ++pp,scanf("%d%d",&Q[pp].l,&Q[pp].r);
        Q[pp].tim=cc,Q[pp].id=pp;
        }
    }
    sort(b+1,b+tot+1);
    tot=unique(b+1,b+tot+1)-b-1;
    for(int i=1;i<=tot;++i) M[b[i]]=i;
    for(int i=1;i<=n;++i) a[i]=M[a[i]];
    for(int i=1;i<=cc;++i) C[i].x=M[C[i].x],C[i].y=M[C[i].y];
    sort(Q+1,Q+pp+1);
    for(int i=1;i<=pp;++i) {
        while(t<Q[i].tim) going(t+1,1),++t;
        while(t>Q[i].tim) going(t,-1),--t;

        while(l<Q[i].l) modify(a[l],-1),++l;
        while(l>Q[i].l) modify(a[l-1],1),--l;
        while(r<Q[i].r) modify(a[r+1],1),++r;
        while(r>Q[i].r) modify(a[r],-1),--r;
        ans[Q[i].id]=calc();

    }
    for(int i=1;i<=pp;++i) printf("%d\n",ans[i]);
}

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原文地址：https://www.cnblogs.com/mfys/p/8519628.html