题面

题解

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//minamoto#include
    
     #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 go(u) for(int i=head[u],v=e[i].v;i;i=e[i].nx,v=e[i].v)template
      
       inline bool cmax(T&a,const T&b){return a
       
        
         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;inline char getc(){return p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<21,stdin),p1==p2)?EOF:*p1++;}long long read(){    R long long 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 long long 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';}const int N=2e5+5,M=2e7+5;typedef long long ll;struct node;typedef node* ptr;unsigned int aaa=19260817;inline unsigned int rd(){aaa^=aaa>>15,aaa+=aaa<<12,aaa^=aaa>>3;return aaa;}inline void swap(R ptr &s,R ptr &t){R ptr p=s;s=t,t=p;}struct node{    ptr lc,rc;bool t;int v,sz;ll sum;unsigned int pr;    inline ptr upd(){return sz=lc->sz+rc->sz+1,sum=lc->sum+rc->sum+v,this;}    inline ptr init(R int val){return sum=v=val,sz=1,pr=rd(),this;}    inline ptr ppd(){return swap(lc,rc),t^=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(ptr p){return ++pp,*pp=*p,pp;}inline ptr pd(ptr p){    if(p->t){        if(p->lc!=e)p->lc=cl(p->lc),p->lc->ppd();        if(p->rc!=e)p->rc=cl(p->rc),p->rc->ppd();        p->t=0;    }return p;}void split(ptr p,int k,ptr &s,ptr &t){    if(p==e)return s=t=e,void();    if(pd(p)->lc->sz
         
          rc,k-s->lc->sz-1,s->rc,t),s->upd(); else 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(pd(s)->pr
          
           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 x){ ptr s,t; split(rt,k,s,t),rt=merge(merge(s,newnode(x)),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);}void rev(ptr &rt,int l,int r){ ptr s,t,p,q; split(rt,r,s,t),split(s,l-1,p,q); rt=merge(merge(p,q->ppd()),t);}ll query(ptr &rt,int l,int r){ ptr s,t,p,q;ll res; split(rt,r,s,t),split(s,l-1,p,q),res=q->sum; return rt=merge(merge(p,q),t),res;}ll lasans,id,op,l,r,x,k;int main(){// freopen("testdata.in","r",stdin); rt[0]=e,e->lc=e->rc=e,lasans=0; for(int Q=read(),i=1;i<=Q;++i){ id=read(),op=read(),rt[i]=rt[id]; switch(op){ case 1:k=read()^lasans,x=read()^lasans,push(rt[i],k,x);break; case 2:k=read()^lasans,pop(rt[i],k);break; case 3:l=read()^lasans,r=read()^lasans,rev(rt[i],l,r);break; case 4:l=read()^lasans,r=read()^lasans,print(lasans=query(rt[i],l,r));break; } } return Ot(),0;}