NOI2017整数

NOI2017 整数

题意：

​ 让你实现两个操作：

• 1 \(a\) \(b\)：将\(x\)加上整数\(a \cdot 2 ^ b\)，其中 \(a\)为一个整数，\(b\)为一个非负整数
• 2 \(k\)：询问 \(x\)在用二进制表示时，位权为\(2 ^ k\)的位的值（即这一位上的\(1\)代表\(2 ^ k\)）

​ 一百万次操作，$ |a| \leq 10^9,b,k\leq30n$。

题解：

​ 线段树+压位，30位一位，没了

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#define fo(i,l,r) for(int i=l;i<=r;i++)
#define of(i,l,r) for(int i=l;i>=r;i--)
#define fe(i,u) for(int i=head[u];i;i=e[i].next)
using namespace std;
typedef long long ll;
inline void open(const char *s)
{
	#ifndef ONLINE_JUDGE
	char str[20];
	sprintf(str,"in%s.txt",s);
	freopen(str,"r",stdin);
//	sprintf(str,"out%s.txt",s);
//	freopen(str,"w",stdout);
	#endif
}
inline int rd()
{
	static int x,f;
	x=0;f=1;
	char ch=getchar();
	for(;ch<'0'||ch>'9';ch=getchar())if(ch=='-')f=-1;
	for(;ch>='0'&&ch<='9';ch=getchar())x=x*10+ch-'0';
	return f>0?x:-x;
}
const int N=1000010,n=N-2,B=30,S=(1<<B)-1;
int m,rt;
namespace Seg{
#define lson tr[o].ls,l,mid
#define rson tr[o].rs,mid+1,r
#define qlson lson,L,min(mid,R)
#define qrson rson,max(mid+1,L),R
struct tree{
	int ls,rs,val0,val1,tag;//1 or 0 and
	tree(){ls=rs=val0=val1=0;tag=-1;}
}tr[N<<1];int cnt=0;
void build(int &o,int l,int r)
{
	o=++cnt;
	if(l==r)return;
	int mid=(l+r)>>1;
	build(lson);
	build(rson);
}
inline void pushup(int o)
{
	tr[o].val1=tr[tr[o].ls].val1|tr[tr[o].rs].val1;
	tr[o].val0=tr[tr[o].ls].val0&tr[tr[o].rs].val0;
}
inline void pushdown(int o)
{
	if(!~tr[o].tag)return;
	int ls=tr[o].ls,rs=tr[o].rs;
	tr[ls].val0=tr[ls].val1=tr[ls].tag=tr[o].tag;
	tr[rs].val0=tr[rs].val1=tr[rs].tag=tr[o].tag;
	tr[o].tag=-1;
}
int find1(int o,int l,int r,int x)
{
	if(tr[o].val0==S)return -1;
	if(l==r)return l;
	pushdown(o);
	int mid=(l+r)>>1,res=-1;
	if(x<=mid)res=find1(lson,x);
	if(~res)return res;
	return find1(rson,x);
}
int find0(int o,int l,int r,int x)
{
	if(!tr[o].val1)return -1;
	if(l==r)return l;
	pushdown(o);
	int mid=(l+r)>>1,res=-1;
	if(x<=mid)res=find0(lson,x);
	if(~res)return res;
	return find0(rson,x);
}
void updata(int o,int l,int r,int x,int d)
{
	if(l==r)return tr[o].val0+=d,tr[o].val1+=d,void();
	pushdown(o);
	int mid=(l+r)>>1;
	if(x<=mid)updata(lson,x,d);
	else updata(rson,x,d);
	pushup(o);
}
void paint(int o,int l,int r,int L,int R,int d)
{
	if(l==L&&r==R)return tr[o].tag=tr[o].val0=tr[o].val1=d,void();
	pushdown(o);
	int mid=(l+r)>>1;
	if(L<=mid)paint(qlson,d);
	if(R>mid)paint(qrson,d);
	pushup(o);
}
int query(int o,int l,int r,int x)
{
	if(l==r)return tr[o].val0;
	pushdown(o);
	int mid=(l+r)>>1;
	if(x<=mid)return query(lson,x);
	return query(rson,x);
}
}
inline void inc(int p,int x)
{
	static int res,y;
//	printf("%d %d\n",p,x);
	res=Seg::query(rt,0,n,p);
	if(res+x<=S)return Seg::updata(rt,0,n,p,x);
	Seg::updata(rt,0,n,p,x-S-1);
	y=Seg::find1(rt,0,n,p+1);
//	printf("%d\n",y);
	if(y>p+1)Seg::paint(rt,0,n,p+1,y-1,0);
	Seg::updata(rt,0,n,y,1);
}
inline void dec(int p,int x)
{
	static int res,y;
//	printf("%d %d\n",p,x);
	res=Seg::query(rt,0,n,p);
	if(res-x>=0)return Seg::updata(rt,0,n,p,-x);
	Seg::updata(rt,0,n,p,S-x+1);
	y=Seg::find0(rt,0,n,p+1);
//	printf("%d\n",y);
	if(y>p+1)Seg::paint(rt,0,n,p+1,y-1,S);
	Seg::updata(rt,0,n,y,-1);
}
inline void gao1()
{
	static int x,y,p;
	x=rd();y=rd();p=y/B;
	if(x>0){
		inc(p,(x<<(y-p*B))&S);
		inc(p+1,x>>(B-y+p*B));
	}
	else{
		x=-x;
		dec(p,(x<<(y-p*B))&S);
		dec(p+1,x>>(B-y+p*B));
	}
}
inline void gao2()
{
	static int x,p,res;
	x=rd();p=x/B;
	res=Seg::query(rt,0,n,p);
	if(res&(1<<(x-p*B)))puts("1");
	else puts("0");
}
int main()
{
	m=rd();rd();rd();rd();
	Seg::build(rt,0,n);
	while(m--){
		int ty=rd();
		if(ty==1)gao1();
		else gao2();
	}
	return 0;
}