正题

题目链接：https://www.luogu.com.cn/problem/P3332

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题目大意

开始$n$个可以重复的集合，要求支持操作

1. $1lrc:$将$c$加入集合$l\sim r$中
2. $2lrk:$查询$l\sim r$的并集中第$k$大的数

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解题思路

此题考验选手的卡常能力（

维护一个树套树，外面是一个权值线段树，然后每个节点套一个区间线段树。然后每次询问和修改的时候就只用使用该节点的线段树的$l\sim r$段即可。

然后区间线段树需要临时开点（废话

之后卡常的话用固定标记，标记不下传，查询的时候再根据沿路的标记来计算答案即可。

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$code$

#pragma GCC optimize(2)
%:pragma GCC optimize(3)
%:pragma GCC optimize("Ofast")
%:pragma GCC optimize("inline")
%:pragma GCC optimize("-fgcse")
%:pragma GCC optimize("-fgcse-lm")
%:pragma GCC optimize("-fipa-sra")
%:pragma GCC optimize("-ftree-pre")
%:pragma GCC optimize("-ftree-vrp")
%:pragma GCC optimize("-fpeephole2")
%:pragma GCC optimize("-ffast-math")
%:pragma GCC optimize("-fsched-spec")
%:pragma GCC optimize("unroll-loops")
%:pragma GCC optimize("-falign-jumps")
%:pragma GCC optimize("-falign-loops")
%:pragma GCC optimize("-falign-labels")
%:pragma GCC optimize("-fdevirtualize")
%:pragma GCC optimize("-fcaller-saves")
%:pragma GCC optimize("-fcrossjumping")
%:pragma GCC optimize("-fthread-jumps")
%:pragma GCC optimize("-funroll-loops")
%:pragma GCC optimize("-fwhole-program")
%:pragma GCC optimize("-freorder-blocks")
%:pragma GCC optimize("-fschedule-insns")
%:pragma GCC optimize("inline-functions")
%:pragma GCC optimize("-ftree-tail-merge")
%:pragma GCC optimize("-fschedule-insns2")
%:pragma GCC optimize("-fstrict-aliasing")
%:pragma GCC optimize("-fstrict-overflow")
%:pragma GCC optimize("-falign-functions")
%:pragma GCC optimize("-fcse-skip-blocks")
%:pragma GCC optimize("-fcse-follow-jumps")
%:pragma GCC optimize("-fsched-interblock")
%:pragma GCC optimize("-fpartial-inlining")
%:pragma GCC optimize("no-stack-protector")
%:pragma GCC optimize("-freorder-functions")
%:pragma GCC optimize("-findirect-inlining")
%:pragma GCC optimize("-fhoist-adjacent-loads")
%:pragma GCC optimize("-frerun-cse-after-loop")
%:pragma GCC optimize("inline-small-functions")
%:pragma GCC optimize("-finline-small-functions")
%:pragma GCC optimize("-ftree-switch-conversion")
%:pragma GCC optimize("-foptimize-sibling-calls")
%:pragma GCC optimize("-fexpensive-optimizations")
%:pragma GCC optimize("-funsafe-loop-optimizations")
%:pragma GCC optimize("inline-functions-called-once")
%:pragma GCC optimize("-fdelete-null-pointer-checks")
#include<cstdio>
#include<cstring>
#include<algorithm>
#define ll long long
#define rint register int
using namespace std;
const int M=5e4+10,N=(5e4)*400+10;
int n,m,tot,cnt,b[M],op[M],l[M],r[M],a[M];
int ls[N],rs[N],lazy[N],root[M*2];
ll val[N];
char buf[1<<23],*head=buf;//fread优化
inline int read(){int x=0,y=0;char ch=getchar();while(ch<'0'||ch>'9'){if(ch=='-')y=1;ch=getchar();}while(ch>='0'&&ch<='9')x=(x<<1)+(x<<3)+(ch^48),ch=getchar();return y?-x:x;
}
template<typename T>
inline T read(){T x=0;int y=0;char ch=getchar();while(ch<'0'||ch>'9'){if(ch=='-')y=1;ch=getchar();}while(ch>='0'&&ch<='9')x=(x<<1)+(x<<3)+(ch^48),ch=getchar();return y?-x:x;
}
void print(int x){if(x<0)putchar('-'),x=-x;if(x>9)print(x/10);putchar(x%10+48);
}
inline void updata(int &x,int l,int r,int L=1,int R=n){if(!x)x=++cnt;val[x]+=(ll)(min(R,r)-max(L,l)+1);if(l<=L&&R<=r){lazy[x]++;return;}rint mid=(L+R)>>1;if(mid>=l)updata(ls[x],l,r,L,mid);if(r>mid)updata(rs[x],l,r,mid+1,R);return;
}
inline ll ask(int &x,int l,int r,int L=1,int R=n,ll t=0){if(!x) return (ll)(min(R,r)-max(L,l)+1)*t;if(l<=L&&R<=r)return val[x]+1ll*(min(R,r)-max(L,l)+1)*t;rint mid=(L+R)>>1;ll ans=0;if(mid>=l)ans+=ask(ls[x],l,r,L,mid,t+lazy[x]);if(r>mid)ans+=ask(rs[x],l,r,mid+1,R,t+lazy[x]);return ans;
}
inline void change(int l,int r,int pos,int L=1,int R=tot,int x=1){updata(root[x],l,r);if(L==R)return;rint mid=(L+R)>>1;if(pos<=mid)change(l,r,pos,L,mid,x<<1);else change(l,r,pos,mid+1,R,x<<1|1);return;
}
inline int query(int l,int r,ll k,int L=1,int R=tot,int x=1){if(L==R)return b[L];rint mid=(L+R)>>1;ll z=ask(root[x<<1|1],l,r);if(z<k)return query(l,r,k-z,L,mid,x<<1);return query(l,r,k,mid+1,R,x<<1|1);
}
int main()
{freopen("P3332_6.in","r",stdin);freopen("P3332_6.ans","w",stdout); n=read();m=read();for(rint i=1;i<=m;i++){op[i]=read();l[i]=read();r[i]=read();a[i]=read();if(op[i]==1)b[++tot]=a[i];}sort(b+1,b+1+tot);tot=unique(b+1,b+1+tot)-b-1;for(rint i=1;i<=m;i++){if(op[i]==1)a[i]=lower_bound(b+1,b+1+tot,a[i])-b,change(l[i],r[i],a[i]);if(op[i]==2)print(query(l[i],r[i],a[i])),putchar('\n');}return 0;
}