#include <cstdio>#include <cstring>#include <cmath>#include <string>#include <iostream>#include <algorithm>#include <queue>#include <set>#include <map>using namespace std;typedef long long ll;#define rep(i, n) for(int i=0; i<(n); ++i)#define for1(i,a,n) for(int i=(a);i<=(n);++i)#define for2(i,a,n) for(int i=(a);i<(n);++i)#define for3(i,a,n) for(int i=(a);i>=(n);--i)#define for4(i,a,n) for(int i=(a);i>(n);--i)#define CC(i,a) memset(i,a,sizeof(i))#define read(a) a=getint()#define print(a) printf("%d", a)#define dbg(x) cout << (#x) << " = " << (x) << endl#define error(x) (!(x)?puts("error"):0)#define rdm(x, i) for(int i=ihead[x]; i; i=e[i].next)inline const int getint() { int r=0, k=1; char c=getchar(); for(; c<‘0‘||c>‘9‘; c=getchar()) if(c==‘-‘) k=-1; for(; c>=‘0‘&&c<=‘9‘; c=getchar()) r=r*10+c-‘0‘; return k*r; }struct node *null;struct node {	node *f, *c[2];	int s, k;	bool rev;	node(int _k=0) { f=c[0]=c[1]=null; k=_k; s=1; rev=0; }	bool d() { return f->c[1]==this; }	void setc(node *x, bool d) { c[d]=x; x->f=this; }	void pushup() { s=1+c[0]->s+c[1]->s; } 	void upd() { if(this==null) return; rev=!rev; swap(c[0], c[1]); }	void pushdown() { if(rev) rev=0, c[0]->upd(), c[1]->upd(); }}*root;void rot(node *x) {	node *f=x->f;	f->pushdown(); x->pushdown(); bool d=x->d();	f->f->setc(x, f->d());	f->setc(x->c[!d], d);	x->setc(f, !d);	f->pushup();	if(f==root) root=x;}void splay(node *x, node *f=null) {	x->pushdown();	while(x->f!=f) 		if(x->f->f==f) rot(x);		else x->d()==x->f->d()?(rot(x->f), rot(x)):(rot(x), rot(x));	x->pushup();}node *sel(node *x, int k) {	if(x==null) return null;	x->pushdown();	int s=x->c[0]->s;	if(s==k) return x;	if(s<k) return sel(x->c[1], k-s-1); return sel(x->c[0], k);}node *getrange(int l, int r) {	splay(sel(root, l-1));	splay(sel(root, r+1), root);	return root->c[1]->c[0];}void build(int l, int r, node *&x) {	if(l>r) return;	int mid=(l+r)>>1;	x=new node(mid);	if(l==r) return;	build(l, mid-1, x->c[0]);	build(mid+1, r, x->c[1]);	if(l<=mid-1) x->c[0]->f=x;	if(mid+1<=r) x->c[1]->f=x;	x->pushup();}int n, m;void init() {	null=new node; null->s=0;	null->f=null->c[0]=null->c[1]=null;	root=new node;	root->setc(new node, 1);	node *x;	build(1, n, x);	root->c[1]->setc(x, 0);	root->c[1]->pushup();	root->pushup();}void rev() {	int l=getint(), r=getint();	node *x=getrange(l, r);	x->upd();}void Pr(node *x) { if(x==null) return; x->pushdown(); Pr(x->c[0]); printf("%d ", x->k); Pr(x->c[1]); }void P() {	node *x=getrange(1, n);	Pr(x);}int main() {	read(n); int m=getint();	init();	while(m--) rev();	P();	return 0;}

第一行为n,m n表示初始序列有n个数，这个序列依次是(1,2……n-1,n)  m表示翻转操作次数
接下来m行每行两个数[l,r] 数据保证 1<=l<=r<=n 

输出一行n个数字，表示原始序列经过m次变换后的结果 

N,M<=100000

平衡树