老早之前就想写了，人生中第一道大分块，调了一上午+下午一个小时，对拍了不知道多少万组，终于过了。
\[\Huge{本题不卡常本题不卡常本题不卡常本题不卡常本题不卡常本题不卡常本题不卡常} \]\[\Huge{快来写快来写快来写快来写快来写快来写快来写快来写快来写快来写快来写} \]solution
由于是Ynoi，考虑分块。
数据范围\(10^5\)，加上这么诡异的操作，先想分块。
区间rank让人以为是[Ynoi2017] 由乃打扑克这道题，但是如果按照由乃打扑克的思路这题的操作就做不了了。
分块是有自己复杂度的就是\(O(\sqrt{n})\)而不是\(O(\log n)\)这意味着分块其实和log的数据结构以及二分法并不是很搭(因为分块的结构本质上就不支持二分)如果我们需要强行嵌入\(log\)的数据结构的话在绝大部分情况下都会使复杂度凭空多出个\(log\)来,这在强调常数的根号算法中绝对是致命的。——来自shadowice1984的题解。
所以可以考虑时间复杂度同为\(O(\sqrt{n})\)的块状数组，由于\(V\le 10^5\)，考虑对值域分块，具体的，就是记\(s1_{i,j}\)表示前\(i\)个序列块中有多少个数在\(j\)值域块中，\(s2_{i,j}\)表示前\(i\)块中有多少个\(j\)，查询时先遍历值域块，确定答案所在值域块的位置，然后再遍历值域块找到答案。
简单的查询就处理完了，那么这个逆天的修改如何处理，有一个经典套路就是用并查集来维护，具体的，记\(rt_{i,x}\)表示第\(i\)块中值为\(x\)的数的代表元，这里我用的是每块中第\(1\)个数的位置。然后用并查集维护位置就可以了，整块就直接令\(rt_{i,y}=\min\{rt_{i,y},rt_{i,x}\}\)即可，考虑散块，仍然考虑暴力修，但是如果每个都修爆炸，只需要重构\(rt_{i,x},rt_{i,y}\)为根的子树就行。
时间复杂度\(O((n+m)\sqrt{n})\)，实现优秀，cin/cout关了同步流都能过！！！
挂一个你谷第2优解。
点此查看代码
#include<bits/stdc++.h>
using namespace std;
#define rep(i,s,t,p) for(int i = s;i <= t; i += p)
#define drep(i,s,t,p) for(int i = s;i >= t; i -= p)
#ifdef LOCAL
FILE *InFile = freopen("in.in","r",stdin),*OutFile = freopen("out.out","w",stdout);
FILE *ErrFile = freopen("err.err","w",stderr);
#else
FILE *InFile = stdin,*OutFile = stdout;
#endif
using ll=long long;using ull=unsigned long long;
using db = double;using ldb = long double;
namespace IO{
const int size = 1<<24;
struct IO{
char buf[size],*p1,*p2,obuf[size],*op = obuf;
#define gc() ((p1==p2&&(p2=(p1=buf)+fread(buf,1,size,stdin),p1==p2))?EOF:*p1++)
inline void pc(char x){if(__builtin_expect(op - obuf == size,1)) fwrite(obuf,1,size,stdout);*op++ = x;}
~IO(){fwrite(obuf,1,op-obuf,stdout);}
template<class T>inline void read(T &x){x = 0;char s = gc();for(;s < '0' || '9' < s;s = gc());for(;'0' <= s && s <= '9';s = gc())x = (x<<1)+(x<<3)+(s^48);}
template<class T,class... Args>inline void read(T &x,Args&... argc){read(x);read(argc...);}
template<class T>
inline void write(T x){static int sta[30],top = 0;do sta[++top] = x%10;while(x /= 10);while(top) pc(sta[top--]+'0');}
inline void write(char x){pc(x);}
template<class T,class... Args>inline void write(T x,Args... argc){write(x);write(argc...);}
}io;
#define read io.read
#define write io.write
}using IO::io;
/*
诶不是，我要处理啥啊。
对于值域块处理1~i中有多少个，然后还要处理每个数。
*/
const int N = 1e5 + 1,SN = N/310,V = 1e5 + 1,SV = N/310,RV = 1e5,cjs = 123;
int n,m,a[N];
int L1[SN],R1[SN],p1[N],lens,sizs;
int L2[SV],R2[SV],p2[V],lenv,sizv;
int s1[SN][SV];//前i块中有多少个数在j块中
int s2[SN][V];//前i块中有多少个j
int s3[SN][V];//第i块中有多少个j
int rt[SN][V],fa[N];
int ct1[SV],ct2[V];//查询时用来记录散块
inline int get_fa(int x){while(x^fa[x]) x = fa[x] = fa[fa[x]];return x;}
inline void Merge(int x,int y){if(x > y) swap(x,y);fa[y] = x;}
//将y并到x中，顺序有关。
inline void upd(int l,int r,int x,int y){
int p = p1[l],q = p1[r],px = p2[x],py = p2[y],sum = 0;
auto rbuild = [&](int p,int l,int r){
if(!s3[p][x]) return 0;
int res = 0,ff = rt[p][x];
vector<int> sonx,sony;
rep(i,L1[p],R1[p],1){
if(get_fa(i) == rt[p][x]){
if(i < l || i > r) sonx.emplace_back(i);
else sony.emplace_back(i),++res;
}
if(fa[i] == rt[p][y]) sony.emplace_back(i);
}
rt[p][x] = rt[p][y] = 0;
for(auto i:sonx){
if(!rt[p][x]) rt[p][x] = fa[i] = i;
else fa[i] = rt[p][x];
a[i] = x;
}
for(auto i:sony){
if(!rt[p][y]) rt[p][y] = fa[i] = i;
else fa[i] = rt[p][y];
a[i] = y;
}
s3[p][x] -= res,s3[p][y] += res;sum += res;
return res;
};
if(p == q){
int res = rbuild(p,l,r);
if(!res) return;
rep(i,p,sizs,1){
s1[i][px] -= res;s1[i][py] += res;
s2[i][x] -= res,s2[i][y] += res;
}
return;
}
rbuild(p,l,R1[p]);
s1[p][px] -= sum;s1[p][py] += sum;
s2[p][x] -= sum,s2[p][y] += sum;
rep(i,p+1,q-1,1){
if(s3[i][x]){
if(!rt[i][y]) rt[i][y] = rt[i][x],a[rt[i][x]] = y;
else Merge(rt[i][y],rt[i][x]);
fa[rt[i][y] = get_fa(rt[i][y])] = rt[i][y];rt[i][x] = 0;
a[get_fa(rt[i][y])] = y;
sum += s3[i][x],s3[i][y] += s3[i][x],s3[i][x] = 0;
}
s1[i][px] -= sum,s1[i][py] += sum;
s2[i][x] -= sum,s2[i][y] += sum;
}
rbuild(q,L1[q],r);
s1[q][px] -= sum;s1[q][py] += sum;
s2[q][x] -= sum,s2[q][y] += sum;
if(!sum) return;
rep(i,q+1,sizs,1){
s1[i][px] -= sum;s1[i][py] += sum;
s2[i][x] -= sum,s2[i][y] += sum;
}
}
inline int qry(int l,int r,int k){
int p = p1[l],q = p1[r];
if(p == q){
vector<int> res;
rep(i,l,r,1) res.emplace_back(a[get_fa(i)]);
nth_element(res.begin(),res.begin()+k-1,res.end());
return *(res.begin()+k-1);
}
rep(i,l,R1[p],1) ++ct2[a[get_fa(i)]],++ct1[p2[a[get_fa(i)]]];
rep(i,L1[q],r,1) ++ct2[a[get_fa(i)]],++ct1[p2[a[get_fa(i)]]];
int sum = 0,pos = 0,ans = 0;
rep(i,1,sizv,1){
sum += ct1[i] + s1[q-1][i] - s1[p][i];
if(sum >= k){sum -= ct1[i] + s1[q-1][i] - s1[p][i];pos = i;break;}
}
rep(j,L2[pos],R2[pos],1){
if(sum + ct2[j] + s2[q-1][j] - s2[p][j] >= k){ans = j;break;}
sum += ct2[j] + s2[q-1][j] - s2[p][j];
}
rep(i,l,R1[p],1) ct2[a[get_fa(i)]]--,ct1[p2[a[get_fa(i)]]]--;
rep(i,L1[q],r,1) ct2[a[get_fa(i)]]--,ct1[p2[a[get_fa(i)]]]--;
return ans;
}
inline void solve(){
read(n,m);rep(i,1,n,1) read(a[i]),fa[i] = i;
lens = 500,sizs = n/lens;
rep(i,1,sizs,1) L1[i] = R1[i-1]+1,R1[i] = i*lens;
if(R1[sizs] < n) sizs++,L1[sizs] = R1[sizs-1]+1,R1[sizs] = n;
lenv = sqrt(RV),sizv = RV/lenv;
rep(i,1,sizv,1) L2[i] = R2[i-1]+1,R2[i] = i*lenv;
if(R2[sizv] < RV) sizv++,L2[sizv] = R2[sizv-1]+1,R2[sizv] = RV;
rep(i,1,sizv,1) rep(j,L2[i],R2[i],1) p2[j] = i;
rep(i,1,sizs,1){
rep(j,L1[i],R1[i],1){
p1[j] = i;
s1[i][p2[a[j]]]++;s2[i][a[j]]++;s3[i][a[j]]++;
if(!rt[i][a[j]]) rt[i][a[j]] = j;
else fa[j] = rt[i][a[j]];
}
rep(j,1,sizv,1) s1[i][j] += s1[i-1][j];
rep(j,1,RV,1) s2[i][j] += s2[i-1][j];
}
rep(test,1,m,1){
int op,l,r,x,y;read(op);
if(op ^ 2) read(l,r,x,y),(x^y)?upd(l,r,x,y):void();
else read(l,r,x),write(qry(l,r,x),'\n');
}
}
signed main(){solve();}