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Give you a series of operations
1: l,val,rl,val,rl,val,r, add a val existing in [l,r]val existing in [l,r]val existing in [l,r] interval.
2. index,kindex,kindex,k ask you the kkk decimal of indexindexindex, if there is no output - 1;
Idea: we can sort the interval of operation 1 by splitting the two endpoints and the time of operation 2 in an array (small is preferred). Mark whether this is a left or right endpoint or a query with opt.
Then we discretize all the values, (Val in operation 1, index in operation 2).
Then iterate through all the ordered operations. If it is currently the left endpoint, add this val to the tree array. Delete if it is the right endpoint. If you want to inquire about the operation, judge whether it exists first, and if it exists, search for the location record answer.
See code for details

#include<bits/stdc++.h>

#define LL long long
#define fi first
#define se second
#define mp make_pair
#define pb push_back

using namespace std;

LL gcd(LL a,LL b){return b?gcd(b,a%b):a;}
LL lcm(LL a,LL b){return a/gcd(a,b)*b;}
LL powmod(LL a,LL b,LL MOD){LL ans=1;while(b){if(b%2)ans=ans*a%MOD;a=a*a%MOD;b/=2;}return ans;}
const int N = 2e5 +11;
int t,n;
struct uzi
{
	int opt,x,val,id;
	bool operator <(const uzi & t)const{
		if(x==t.x)return id<t.id;
		return x<t.x;
	}
}p[N];
int dx=0,dy=0,dz=0;
int c[N],a[N],ans[N],len;
int add(int now,int x){
	for(int i=now;i<=len;i+=(i&(-i)))c[i]+=x;
}
int que(int now){
	int sum=0;
	for(int i=now;i;i-=(i&(-i)))sum+=c[i];
	return sum;
}
int get(int sum){
	int l=1,r=len,ans=-1;
	while(l<=r){
		int mid=l+r>>1;
		if(que(mid)>=sum)ans=mid,r=mid-1;
		else l=mid+1;
	}
	return ans;
}
int now;
int main(){
	for(scanf("%d",&t);t;t--){
		scanf("%d",&n);
		dx=dy=dz=0;
		memset(c,0,sizeof c);
		memset(ans,0,sizeof ans);
		for(int i=1;i<=n;i++){
			int opt,x,y,z;
			scanf("%d%d%d",&opt,&x,&y);
			a[++dz]=y;
			if(opt==1){
				scanf("%d",&z);
				p[++dx]={1,x,y,i};
				p[++dx]={-1,z+1,y,i};
			}else p[++dx]={2,x,y,i};
		}
		sort(a+1,a+1+dz);
		len=unique(a+1,a+1+dz)-a-1;
		for(int i=1;i<=dx;i++)p[i].val=lower_bound(a+1,a+1+len,p[i].val)-a;
		sort(p+1,p+1+dx);
		for(int i=1;i<=dx;i++){
			if(p[i].opt!=2){
				add(p[i].val,p[i].opt);
			}else{
				if(que(len)<p[i].val)ans[i]=-1;
				else ans[i]=get(p[i].val);
			}
		}
		printf("Case %d:\n",++now );
		for(int i=1;i<=dx;i++){
			if(!ans[i])continue;
			if(ans[i]==-1)printf("-1\n");
			else printf("%d\n",a[ans[i]] );
		}
	}
	return 0;
}

An online version of the chairman tree was added.
Maintain the right endpoint with priority queue, and update the disappeared val when the query is needed. Then directly query the k of the chairman tree.

#include <bits/stdc++.h>
#define LL long long
#define fi first
#define se second
#define mp make_pair
#define pb push_back

using namespace std;

LL gcd(LL a,LL b){return b?gcd(b,a%b):a;}
LL lcm(LL a,LL b){return a/gcd(a,b)*b;}
LL powmod(LL a,LL b,LL MOD){LL ans=1;while(b){if(b%2)ans=ans*a%MOD;a=a*a%MOD;b/=2;}return ans;}
const int N = 1e5 +11;
const int M =1e6;
int n,m;
struct uzi
{
	int sum,l,r;
}T[N*40];
int now,a[N],b[N],t[N];
void update(int l,int r,int &x,int y,int pos,int val){
	x=++now;
	T[++x]=T[y];T[x].sum+=val;
	if(l==r)return ;
	int mid=l+r>>1;
	if(mid>=pos)update(l,mid,T[x].l,T[y].l,pos,val);
	else update(mid+1,r,T[x].r,T[y].r,pos,val);
}
int get(int now,int l,int r,int k){
	if(T[now].sum<k)return -1;
	if(l==r)return l;
	int mid=l+r>>1;
	if(T[T[now].l].sum>=k)return get(T[now].l,l,mid,k);
	else return get(T[now].r,mid+1,r,k-T[T[now].l].sum);
}
int ab,noa;
int main(){
	for(scanf("%d",&ab);ab;ab--){
		printf("Case %d:\n",++noa );
		scanf("%d",&n);
		int cnt=0;
		now=0;
		priority_queue<pair<int,int> >q;
		for(int i=1;i<=n;i++){
			int opt,l,r,z,val;
			scanf("%d%d%d",&opt,&l,&val);
			if(opt==1){
				scanf("%d",&r);
				update(1,M+1,cnt,cnt,val,1);
				q.push({-r,val});
			}else{
				while(!q.empty()&&-q.top().fi<l){
					update(1,M+1,cnt,cnt,q.top().se,-1);
					q.pop();
				}
				printf("%d\n",get(cnt,1,M+1,val));
			}
		}
	}
	return 0;
}