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:
You are given two integers l and r. Let ci be the number of occurrences of i in al: r, where al: r is the subarray of a from l-th element to r-th inclusive. Find the Mex of {c0, c1, …, c109}
You are given two integers p to x. Change ap 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, …, an (1 ≤ ai ≤ 109).
Each of the next q lines describes a single query.
The first type of query is described by three integers ti = 1, li, ri, where 1 ≤ li ≤ ri ≤ n — the bounds of the subarray.
The second type of query is described by three integers ti = 2, pi, xi, where 1 ≤ pi ≤ n is the index of the element, which must be changed and 1 ≤ xi ≤ 109 is the new value.
Output
For each query of the first type output a single integer — the Mex of {c0, c1, …, c109}.
Example
Input
10 4
1 2 3 1 1 2 2 2 9 9
1 1 1
1 2 8
2 7 1
1 2 8
Output
2
3
2
Note
The subarray of the first query consists of the single element — 1.
The subarray of the second query consists of four 2s, one 3 and two 1s.
The subarray of the fourth query consists of three 1s, three 2s and one 3.
题意：
给你n个数，有两个操作，1 l r表示在l到r这段区间，询问{c0,c1,c2,⋯,c109}的Mex，而ci表示数值i在l r中的出现次数，Mex代表未出现的最小的正整数。举个例子:1 2 3 4 3 2 2 ，最小为4，1 2 3 3 2 1最小是1.。。。2 p x 代表把a[p]改成x
题解：
带修莫队，cnt1[x]维护x出现的次数cnt2[x]维护出现x次的数有多少个。把所有2的操作记录下来，在记录tim代表这个1操作之前有多少2操作，如果在莫队的时候的t小了，那就增加2操作，反之减少
#include<bits/stdc++.h>
using namespace std;
const int N=1e5+5;
int len;
struct node
{
int l,r,t,bl,br,indx;
node(){}
node(int l,int r,int t,int indx):l(l),r(r),t(t),indx(indx){
bl=l/len,br=r/len;
}
bool operator< (const node& a)const
{
if(bl==a.bl&&br==a.br)
return t<a.t;
if(bl==a.bl)
return br<a.br;
return bl<a.bl;
}
}que[N];
struct query
{
int pos,pre,val;// 位置，上一个数，当前的数
query(){}
query(int pos,int pre,int val):pos(pos),pre(pre),val(val){}
}c[N];
int cnt1[2*N],cnt2[2*N],a[N],b[N*2],now[N],cnt,ans[N];
void add(int pos)
{
cnt2[cnt1[pos]]--;
cnt1[pos]++;
cnt2[cnt1[pos]]++;
}
void del(int pos)
{
cnt2[cnt1[pos]]--;
cnt1[pos]--;
cnt2[cnt1[pos]]++;
}
int ll,rr;
void change(int pos,int x)
{
if(pos>=ll&&pos<=rr){
del(now[pos]);
add(x);
}
now[pos]=x;
}
int main()
{
int n,q;
scanf("%d%d",&n,&q);
len=pow(n,0.666667);
for(int i=1;i<=n;i++)
{
scanf("%d",&a[i]);
now[i]=a[i];
b[++cnt]=a[i];
}
/*for(int i=1;i<=n;i++)
cout<<now[i]<<" ";
cout<<endl;*/
int op,qnum=0,tim=0;
for(int i=1;i<=q;i++)
{
scanf("%d%d%d",&op,&ll,&rr);
if(op==1)
que[++qnum]=node(ll,rr,tim,qnum);
else
b[++cnt]=rr,c[++tim]=query(ll,now[ll],rr),now[ll]=rr;
}
/*for(int i=1;i<=n;i++)
cout<<now[i]<<" ";
cout<<endl;*/
sort(b+1,b+1+cnt);
sort(que+1,que+qnum+1);
int all=unique(b+1,b+1+cnt)-b-1;
for(int i=1;i<=n;i++)
now[i]=lower_bound(b+1,b+1+all,a[i])-b;
for(int i=1;i<=tim;i++)
{
c[i].pre=lower_bound(b+1,b+1+all,c[i].pre)-b;
c[i].val=lower_bound(b+1,b+1+all,c[i].val)-b;
}
ll=1,rr=0;
int t=0;
for(int i=1;i<=qnum;i++)
{
while(ll>que[i].l)
ll--,add(now[ll]);
while(rr<que[i].r)
rr++,add(now[rr]);
while(ll<que[i].l)
del(now[ll]),ll++;
while(rr>que[i].r)
del(now[rr]),rr--;
while(t<que[i].t)
t++,change(c[t].pos,c[t].val);
while(t>que[i].t)
change(c[t].pos,c[t].pre),t--;
int mex=1;
while(cnt2[mex])
mex++;
ans[que[i].indx]=mex;
}
for(int i=1;i<=qnum;i++)
printf("%d\n",ans[i]);
return 0;
}