根据目前的测试结果，第一个点先有大约5e5个insert操作，然后是5e5个delete操作，最后一个是4操作

前面insert运行速度还可以，但后面Delete明显很慢，本机大概20s+出答案，答案是正确的

提交记录与代码

#include<iostream>
#include<cstdio>
#include<ctime>
#include<cstdlib>
#define ll long long
int n,m;
ll lst,ans;
class Treap{
	public:
	int root,tot;
	class node{
		public:
		int l,r,size;
		ll v;
	}a[1100010];
	Treap(){
		root=tot=0;
	}void up(int x){
		if(x!=0)a[x].size=a[a[x].l].size+a[a[x].r].size+1;
	}int newnode(ll val){
		a[++tot].v=val;
		a[tot].size=1;
		return tot;
	}void split(int p,ll val,int &x,int &y){
		if(p==0)x=y=0;
		else if(a[p].v<=val)x=p,split(a[p].r,val,a[p].r,y);
		else y=p,split(a[p].l,val,x,a[p].l);
		up(p);
		return;
	}int merge(int x,int y){
		if(!x||!y)return x+y;
		int rd=std::rand()%(a[x].size+a[y].size);
		int l=a[y].l,r=a[x].r;
		if(rd<a[x].size){
			a[x].r=y;
			a[y].l=merge(r,l);
			up(y),up(x);
			return x;
		}else {
			a[y].l=x;
			a[x].r=merge(r,l);
			up(x),up(y);
			return y;
		}
	}void insert(ll val){
		int x,y;
		split(root,val,x,y);
		root=merge(merge(x,newnode(val)),y);
		return;
	}void Delete(ll val){
		int x,y,z;
		split(root,val,x,z);
		split(x,val-1,x,y);
		y=merge(a[y].l,a[y].r);
		root=merge(merge(x,y),z);
		return;
	}int rank(ll val){
		int x,y,rk;
		split(root,val-1,x,y);
		rk=a[x].size;
		root=merge(x,y);
		return rk;
	}int rerank(int x,int rk){
		while(x){
			if(a[a[x].l].size>rk)x=a[x].l;
			else if(a[a[x].l].size==rk)break;
			else rk-=a[a[x].l].size+1,x=a[x].r;
		}return x;
	}int fpre(ll val){
		int x,y,an;
		split(root,val-1,x,y);
		an=rerank(x,a[x].size-1);
		root=merge(x,y);
		return an;
	}int fnxt(ll val){
		int x,y,an;
		split(root,val,x,y);
		an=rerank(y,0);
		root=merge(x,y);
		return an;
	}
}bt;
int main(){
	std::srand(time(NULL));
	scanf("%d%d",&n,&m);
	for(int i=1;i<=n;i++){
		ll val;
		scanf("%lld",&val);
		bt.insert(val);
	}
	for(int i=1;i<=m;i++){
		ll opt,x;
		scanf("%lld%lld",&opt,&x);
		x^=lst;
		if(opt==1)bt.insert(x);
		if(opt==2)bt.Delete(x);
		if(opt==3)lst=bt.rank(x)+1;
		if(opt==4)lst=bt.a[bt.rerank(bt.root,x-1)].v;
		if(opt==5)lst=bt.a[bt.fpre(x)].v;
		if(opt==6)lst=bt.a[bt.fnxt(x)].v;
		if(opt>=3)ans^=lst;
	}printf("%lld",ans);
	return 0;
}