Tree Construction
During the programming classes Vasya was assigned a difficult problem. However, he doesn't know how to code and was unable to find the solution in the Internet, so he asks you to help.
You are given a sequence a, consisting of n distinct integers, that is used to construct the binary search tree. Below is the formal description of the construction process.
- First element a1 becomes the root of the tree.
- Elements a2, a3, ..., an are added one by one. To add element ai one needs to traverse the tree starting from the root and using the following rules:
- The pointer to the current node is set to the root.
- If ai is greater than the value in the current node, then its right child becomes the current node. Otherwise, the left child of the current node becomes the new current node.
- If at some point there is no required child, the new node is created, it is assigned value ai and becomes the corresponding child of the current node.
Input
The first line of the input contains a single integer n (2 ≤ n ≤ 100 000) — the length of the sequence a.
The second line contains n distinct integers ai (1 ≤ ai ≤ 109) — the sequence aitself.
Output
Output n - 1 integers. For all i > 1 print the value written in the node that is the parent of the node with value ai in it.
Examples
Input
3
1 2 3
Output
1 2
Input
5
4 2 3 1 6
Output
4 2 2 4
Note
Picture below represents the tree obtained in the first sample.
Picture below represents the tree obtained in the second sample.
sol：一个很显然的性质就是当前这个点父亲的权值一定是他的前驱后者后继，但是怎么判断是前驱的右儿子或者后继的左儿子嫩？貌似有点蛋碎，挂了两次之后知道了是出现时间最晚的那个就是了，前驱后继写（copy）一个splay就可以了
#include <bits/stdc++.h> using namespace std; typedef int ll; inline ll read() { ll s=0; bool f=0; char ch=' '; while(!isdigit(ch)) { f|=(ch=='-'); ch=getchar(); } while(isdigit(ch)) { s=(s<<3)+(s<<1)+(ch^48); ch=getchar(); } return (f)?(-s):(s); } #define R(x) x=read() inline void write(ll x) { if(x<0) { putchar('-'); x=-x; } if(x<10) { putchar(x+'0'); return; } write(x/10); putchar((x%10)+'0'); return; } #define W(x) write(x),putchar(' ') #define Wl(x) write(x),putchar(' ') const int N=200005,inf=0x3f3f3f3f; int n,a[N],b[N]; namespace Pht { int Points,Root; int Child[N][2]; int Parent[N]; int Quanzhi[N]; int Cnt[N]; int Size[N]; inline void Init(); inline int Check(int x); inline void PushUp(int x); inline void Rotate(int x); inline void Splay(int At,int To); inline void Insert(int Val); inline void Remove(int Val); inline void Find(int Val); inline int Ask_Lower(int Val); inline int Ask_Upper(int Val); inline int Ask_Kth(int Id); inline void Solve(); inline void Init() { Points=Root=0; Insert(-inf); Insert(inf); } inline int Check(int x) { return (Child[Parent[x]][0]==x)?0:1; } inline void PushUp(int x) { Size[x]=Size[Child[x][0]]+Size[Child[x][1]]+Cnt[x]; } inline void Rotate(int x) { int y,z,oo; y=Parent[x]; z=Parent[y]; oo=Check(x); Child[y][oo]=Child[x][oo^1]; Parent[Child[x][oo^1]]=y; Child[z][Check(y)]=x; Parent[x]=z; Child[x][oo^1]=y; Parent[y]=x; PushUp(x); PushUp(y); } inline void Splay(int At,int To) { while(Parent[At]!=To) { int Father=Parent[At]; if(Parent[Father]==To) { Rotate(At); } else if(Check(At)==Check(Father)) { Rotate(Father); Rotate(At); } else { Rotate(At); Rotate(At); } } if(To==0) Root=At; } inline void Insert(int Val) { int Now=Root,Par=0; while(Now&&(Quanzhi[Now]!=Val)) { Par=Now; Now=Child[Now][(Val>Quanzhi[Now])?1:0]; } if(Now) { Cnt[Now]++; Size[Now]++; } else { Now=++Points; if(Par) { Child[Par][(Val>Quanzhi[Par])?1:0]=Now; } Parent[Now]=Par; Child[Now][0]=Child[Now][1]=0; Cnt[Now]=Size[Now]=1; Quanzhi[Now]=Val; } Splay(Now,0); } inline void Remove(int Val) { int Lower=Ask_Lower(Val),Upper=Ask_Upper(Val); Splay(Lower,0); Splay(Upper,Lower); if(Cnt[Child[Upper][0]]>1) { Cnt[Child[Upper][0]]--; Size[Child[Upper][0]]--; Splay(Child[Upper][0],0); } else { Child[Upper][0]=0; } } inline void Find(int Val) { int Now=Root; if(!Now) return; while(Child[Now][(Val>Quanzhi[Now])?1:0]&&(Quanzhi[Now]!=Val)) { Now=Child[Now][(Val>Quanzhi[Now])?1:0]; } Splay(Now,0); } inline int Ask_Lower(int Val) { Find(Val); int Now=Root; if(Quanzhi[Now]<Val) return Now; Now=Child[Now][0]; while(Child[Now][1]) Now=Child[Now][1]; return Now; } inline int Ask_Upper(int Val) { Find(Val); int Now=Root; if(Quanzhi[Now]>Val) return Now; Now=Child[Now][1]; while(Child[Now][0]) Now=Child[Now][0]; return Now; } inline int Ask_Kth(int Id) { int Now=Root; if(Size[Root]-1<Id) return 0; for(;;) { if(Size[Child[Now][0]]>=Id) { Now=Child[Now][0]; } else if(Size[Child[Now][0]]+Cnt[Now]<Id) { Id-=Size[Child[Now][0]]+Cnt[Now]; Now=Child[Now][1]; } else return Now; } } int Time[N]; inline void Solve() { Init(); int i,Min=a[1],Max=a[1],tt=0; Insert(a[1]); Time[Root]=++tt; for(i=2;i<=n;i++) { int Lower=Ask_Lower(a[i]),Upper=Ask_Upper(a[i]); // printf("T[%d]=%d T[%d]=%d ",Lower,Time[Lower],Upper,Time[Upper]); if(Time[Lower]>Time[Upper]) { W(Quanzhi[Lower]); } else { W(Quanzhi[Upper]); } Insert(a[i]); Time[Root]=++tt; // puts(""); } } } int main() { int i; R(n); for(i=1;i<=n;i++) R(a[i]); Pht::Solve(); return 0; } /* Input 3 1 2 3 Output 1 2 Input 5 4 2 3 1 6 Output 4 2 2 4 Input 10 991309218 517452607 870021923 978357992 136426010 10601767 302627526 883615372 163475700 600546765 Output 991309218 517452607 870021923 517452607 136426010 136426010 978357992 302627526 870021923 */