BZOJ4415: [Shoi2013]发牌

显然可以线段树或树状数组上二分。

然而直接写splay在bzoj上并不会T。

然而发这题的目的只是因为我又忘了return了啊啊啊啊(TдT)

内心十分崩溃。关键是在本地还能过。

#include<cstdio>
#include<algorithm>
#define L(t) t->c[0]
#define R(t) t->c[1]
#define Z(t) (L(t)->s+1)
#define N 700005
#define M (s+t>>1)
using std::swap;
struct node{
	int v,s;
	node *c[2];
}e[N],*back=e+1,
*null=e,*root;
node* update(node* t){
	t->s=R(t)->s+Z(t);
	return t;
}
void link(bool i,
node*& t,node*& s){
	node* d=t->c[i];
	t->c[i]=s;
	s=update(t),t=d;
}
node* splay(int v,
node*& t=root){
	node* d[]={null,null};
	while(v!=Z(t)){
		bool i=v>Z(t);
		v-=i*Z(t);
		if(v!=Z(t->c[i])
		&&i==v>Z(t->c[i])){
			v-=i*Z(t->c[i]);
			link(i,t,
			t->c[i]->c[i^1]);
		}
		link(i,t,d[i]);
	}
	for(int i=0;i!=2;++i){
		node* s=t->c[i^1];
		while(d[i]!=null)
			link(i,d[i],s);
		t->c[i^1]=s;
	}
	return update(t);
}
node* build(int s,int t){
	if(s<=t){
		node* a=back++;
		a->v=M;
		L(a)=build(s,M-1);
		R(a)=build(M+1,t);
		return update(a);
	}
	return null;
}
node*& splay(int s,int t){
	splay(s);
	return L(splay(
	t-s+2,R(root)));
}
struct io_t{
	char p[1<<25],*s;
	char e[1<<25],*t;
	int a[24];
	io_t():s(p),t(e){
		fread(s,1,
		sizeof p,stdin);
	}
	~io_t(){
		fwrite(e,1,
		t-e-1,stdout);
	}
	operator int(){
		static int v;
		v=0;
		while(*s<48)
			++s;
		do
			v=v*10+*s++-48;
		while(*s>32);
		return v;
	}
	void print(int v){
		static int* q=a;
		if(!v)*t++=48;
		else{
			if(v<0)
				*t++=45,
				v*=-1;
			while(v)
				*q++=v%10+48,
				v/=10;
			while(q!=a)
				*t++=*--q;
		}
		*t++=10;
	}
}ip;
int n=ip,s=1;
int main(){
	root=build(0,n+1);
	for(;n;--n){
		s=(s+ip-1)%n+1;
		ip.print(
		splay(s,s)->v);
		L(R(root))=null;
	}
}