# LOJ#120. 持久化序列（FHQ Treap）

## 题解

//minamoto
#include<bits/stdc++.h>
#define R register
#define inline __inline__ __attribute__((always_inline))
#define fp(i,a,b) for(R int i=(a),I=(b)+1;i<I;++i)
#define fd(i,a,b) for(R int i=(a),I=(b)-1;i>I;--i)
template<class T>inline bool cmax(T&a,const T&b){return a<b?a=b,1:0;}
template<class T>inline bool cmin(T&a,const T&b){return a>b?a=b,1:0;}
using namespace std;
char buf[1<<21],*p1=buf,*p2=buf;
R int res,f=1;R char ch;
while((ch=getc())>'9'||ch<'0')(ch=='-')&&(f=-1);
for(res=ch-'0';(ch=getc())>='0'&&ch<='9';res=res*10+ch-'0');
return res*f;
}
char sr[1<<21],z[20];int K=-1,Z=0;
inline void Ot(){fwrite(sr,1,K+1,stdout),K=-1;}
void print(R int x){
if(K>1<<20)Ot();if(x<0)sr[++K]='-',x=-x;
while(z[++Z]=x%10+48,x/=10);
while(sr[++K]=z[Z],--Z);sr[++K]='\n';
}
unsigned int aaa=19260817;
inline unsigned int rd(){aaa^=aaa>>15,aaa+=aaa<<12,aaa^=aaa>>3;return aaa;}
const int N=3e5+5,M=1e7+5;
struct node;typedef node* ptr;
struct node{
ptr lc,rc;int sz,v;unsigned int pr;
inline ptr init(R int val){return sz=1,v=val,pr=rd(),this;}
inline ptr upd(){return sz=lc->sz+rc->sz+1,this;}
}e[M],*rt[N],*pp=e;
inline ptr newnode(R int v){return ++pp,pp->lc=pp->rc=e,pp->init(v);}
inline ptr cl(R ptr p){return ++pp,*pp=*p,pp;}
void split(ptr p,int k,ptr &s,ptr &t){
if(p==e)return s=t=e,void();
if(p->lc->sz<k)return s=cl(p),split(s->rc,k-s->lc->sz-1,s->rc,t),s->upd(),void();
t=cl(p),split(t->lc,k,s,t->lc),t->upd();
}
ptr merge(ptr s,ptr t){
if(s==e)return t;if(t==e)return s;
if(s->pr<t->pr)return s->rc=merge(s->rc,t),s->upd();
return t->lc=merge(s,t->lc),t->upd();
}
void push(ptr &rt,int k,int v){
ptr s,t;
split(rt,k-1,s,t),rt=merge(merge(s,newnode(v)),t);
}
void pop(ptr &rt,int k){
ptr s,t,p,q;
split(rt,k,s,t),split(s,k-1,p,q),rt=merge(p,t);
}
int query(ptr &rt,int k){
ptr s,t,p,q;int now;
split(rt,k,s,t),split(s,k-1,p,q),now=q->v;
return rt=merge(merge(p,q),t),now;
}
int main(){
//	freopen("testdata.in","r",stdin);
//	freopen("testdata.out","w",stdout);
rt[0]=e,e->lc=e->rc=e;