CodeForces 940F Machine Learning(带修改的莫队)

本文链接：https://blog.youkuaiyun.com/dsaghjkye/article/details/82285145

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.

题意给定一个长度为n 的数组a ，并且要求执行q 个操作，有两种不同的操作：

1. 询问区间【l,r】内未出现的最小整数是啥从 1 开始，比如 3个5 1 个 10 ，那么数字是 2

2.更改某一个点上的值

区间问题不要求在线可以考虑莫队因为有时间轴所以要带修改

因为数据范围大但n小所以先离散化

#include <bits/stdc++.h>
const int maxn=2e5+10;
using namespace std;
int a[maxn],now[maxn],be[maxn],num[maxn],cnt[maxn],ans[maxn];//a 代表输入数据对应离散化后数字    now 修改后代表的数字，每当修改时 会把now传入记录之前值和现在值    be 分块值   num   数字的个数    cnt  数字个数的 个数   ans 保存结果
int n,m,l,r,t;
int msz;
map<int,int>M;//离散化 记录对应值
struct Query
{
    int l,r,t,id;
}q[maxn];
int cmp(Query n1,Query n2)//排序
{
    if(be[n1.l]==be[n2.l])
    {
        if(be[n1.r]==be[n2.r])
         return n1.t<n2.t;
         return n1.r<n2.r;
    }
    return n1.l<n2.l;
}
struct nn
{
    int pos,neww,oldd;//位置 ，现在的新值  ，之前的值
}c[maxn];
int compress(int x)
{
    if(!M.count(x)) M[x]=++msz; //如果未出现过 ，赋予新值
    return M[x];
}
void add(int val,int d)
{
    if(num[val]>0) cnt[num[val]]--;
    num[val]+=d;
    if(num[val]>0) cnt[num[val]]++;
}
void go(int idx,int val)
{
    if(l<=idx&&idx<=r) //要修改的值在当前l r 中 那么得先去掉之前的值在添新值，不在则无影响
    {
        add(a[idx],-1);
        add(val,1);
    }
    a[idx]=val; //修改
}
int get() { //得到最小整数
    for(int i = 1; ;++i) {
        if(cnt[i] == 0) return i;
    }
}
int main()
{
    msz=0;
   int x,qsz=0,csz=0,y,op;
   scanf("%d%d",&n,&m);
   int tt=pow(n,2.0/3.0);//
   for(int i=1;i<=n;i++)
   {
    scanf("%d",&x);
    now[i]=a[i]=compress(x);//获得对应值
    be[i]=(i-1)/tt+1;
   }
   for(int i=0;i<m;i++)
   {
       scanf("%d%d%d",&op,&x,&y);
       if(op==1) q[++qsz]=(Query){x,y,csz,qsz};
       else
       {
           y=compress(y);
           c[++csz]=(nn){x,y,now[x]};
           now[x]=y;//之前的值修改
       }
   }
   sort(q+1,q+qsz+1,cmp);
   l=1,r=0,t=0;
   for(int i=1;i<=qsz;i++)
   {
      while(t<q[i].t) go(c[t+1].pos,c[t+1].neww),++t;
      while(t>q[i].t) go(c[t].pos,c[t].oldd),--t;
      while(l<q[i].l) add(a[l++],-1);
      while(l>q[i].l) add(a[--l],1);
      while(r<q[i].r) add(a[++r],1);
      while(r>q[i].r) add(a[r--],-1);
      ans[q[i].id]=get();
   }
   for(int i=1;i<=qsz;i++)
    printf("%d\n",ans[i]);
}