题意：
给一系列数，执行一系列的反转操作，要求最后稳定排序这些数，并且输出每次反转的右边界。
题解：
怎么运用伸展树去解决呢？（反转操作。伸展树的基本操作）
对原始数据记录值和下标，然后按照值优先排序。伸展树存放下标。然后扫描一遍所有的数：
第一个数在x位置，把伸展树值为x的节点伸展到根，左子树size+1即为所求的右边界。然后把x左边的节点反转，删除根（x节点）。
顺次第二个数，第三个~
#include<bits/stdc++.h>using namespace std;int n;
const int maxn = 1e5+99;struct splaytree
{int pre[maxn],ch[maxn][2],sz[maxn],rev[maxn],root;void update(int r)///反转更新{if(r == 0)return;rev[r]^=1;//下面的节点翻转抵消swap(ch[r][0],ch[r][1]);}void pushdown(int r){if(rev[r]){update(ch[r][0]);update(ch[r][1]);rev[r] = 0;}}void pushup(int r){sz[r] = 1 + sz[ch[r][0]] + sz[ch[r][1]];}void rotate(int x,int d){int y = pre[x];pushdown(y);pushdown(x);ch[y][d^1] = ch[x][d];pre[ch[x][d]] = y;if(pre[y]) ch[pre[y]][ch[pre[y]][1]==y]=x;pre[x] = pre[y];ch[x][d] = y;pre[y] = x;pushup(y);}void splay(int r,int goal){pushdown(r);while(pre[r] != goal){if(pre[pre[r]] == goal)rotate(r,ch[pre[r]][0]==r);else{int y = pre[r];int d = ch[pre[y]][0]==y;if(ch[r][d]==r){rotate(r,!d);rotate(r,d);}else{rotate(y,d);rotate(r,d);}}}pushup(r);if(goal == 0)root = r;}void newnode(int &r,int fa,int k){r = k;pre[r] = fa;ch[r][0] = ch[r][1] =0;sz[r] = 1;}void build(int &x,int l,int r,int fa){if(l>r)return;int m = (l+r)>>1;newnode(x,fa,m);build(ch[x][0],l,m-1,x);build(ch[x][1],m+1,r,x);pushup(x);}void init(){root = 0;ch[root][0] = ch[root][1] = rev[root] = sz[root] = pre[root] = 0;build(root,1,n,0);}int getmax(int r){pushdown(r);while(ch[r][1]){r = ch[r][1];pushdown(r);}return r;}void remove()/**删根*/{if(ch[root][0]==0){root = ch[root][1];pre[root] = 0;}else{int m = getmax(ch[root][0]);splay(m,root);ch[m][1] = ch[root][1];pre[ch[root][1]] = m;root = m;pre[m] = 0;pushup(root);}}}st;struct node
{int num,id;bool operator < (const node& a)const{if(num == a.num)return id<a.id;return num<a.num;}
}ns[maxn];int main()
{while(scanf("%d",&n)==1&&n){st.init();for(int i=1;i<=n;i++){scanf("%d",&ns[i].num);ns[i].id = i;}sort(ns+1,ns+1+n);for(int i = 1;i<n;i++){st.splay(ns[i].id,0);st.update(st.ch[st.root][0]);printf("%d ",st.sz[st.ch[st.root][0]] + i);st.remove();}printf("%d\n",n);}
}