题目大意：给定n个操作，每个操作有四种形式，操作之后若<L就变成L，>R就变成R，现在给定q个输入，求他们的输出
n,q<=10W
将这q个数建立线段树，四个操作都可以在线段树上完成
但是溢出怎么办呢？
容易发现若x<=y，那么操作过后x一定<=y
因为四个操作都是线性的，溢出也不会反转两个数的大小关系
那么我们可以预先将q个数排序 那么溢出的数一定是连续的两段 区间修改就行了
至于怎么找两端溢出的区间…… 每个节点维护一个区间最大值和最小值就好了
此外数据还是比较良心的 不会出现爆long long这种P事 放心写吧
这么简单的思路我他妈的居然写了整整四遍的代码- - 下次写代码之前一定要想清楚- - 至少TM先手模拟过样例啊QAQ
#include <cmath>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#define M 100100
using namespace std;
int n,m,L,R;
int q[M];
long long a[M];
struct Operation{
int type,x;
friend istream& operator >> (istream &_,Operation &o)
{
char p[10];
scanf("%s%d",p,&o.x);
switch(p[0])
{
case '+':o.type=1;break;
case '-':o.type=2;break;
case '*':o.type=3;break;
case '@':o.type=4;break;
}
return _;
}
}operations[M];
struct Segtree{
Segtree *ls,*rs;
long long times_mark,k,b;
long long min_num,max_num;
Segtree():ls(0x0),rs(0x0),times_mark(1),k(0),b(0) {}
void Build_Tree(int x,int y)
{
int mid=x+y>>1;
min_num=a[x];
max_num=a[y];
if(x==y) return ;
(ls=new Segtree)->Build_Tree(x,mid);
(rs=new Segtree)->Build_Tree(mid+1,y);
}
void Get_Mark(int x,int y,long long _,long long __,long long ___)
{
min_num=_*min_num+__*a[x]+___;
max_num=_*max_num+__*a[y]+___;
times_mark*=_;k*=_;b*=_;
k+=__;
b+=___;
}
void Push_Down(int x,int y)
{
int mid=x+y>>1;
ls->Get_Mark(x,mid,times_mark,k,b);
rs->Get_Mark(mid+1,y,times_mark,k,b);
times_mark=1;k=b=0;
}
int Get_Ans(int x,int y,int pos)
{
int mid=x+y>>1;
if(x==y) return max_num;
Push_Down(x,y);
if(pos<=mid)
return ls->Get_Ans(x,mid,pos);
else
return rs->Get_Ans(mid+1,y,pos);
}
void Get_L(int x,int y)
{
int mid=x+y>>1;
if(x==y)
{
Get_Mark(x,y,0,0,L);
return ;
}
Push_Down(x,y);
if(rs->min_num<L)
ls->Get_Mark(x,mid,0,0,L),rs->Get_L(mid+1,y);
else
ls->Get_L(x,mid);
min_num=ls->min_num;
max_num=rs->max_num;
}
void Get_R(int x,int y)
{
int mid=x+y>>1;
if(x==y)
{
Get_Mark(x,y,0,0,R);
return ;
}
Push_Down(x,y);
if(ls->max_num>R)
rs->Get_Mark(mid+1,y,0,0,R),ls->Get_R(x,mid);
else
rs->Get_R(mid+1,y);
min_num=ls->min_num;
max_num=rs->max_num;
}
}*tree=new Segtree;
int main()
{
int i;
cin>>n>>L>>R;
for(i=1;i<=n;i++)
cin>>operations[i];
cin>>m;
for(i=1;i<=m;i++)
scanf("%d",&q[i]),a[i]=q[i];
sort(a+1,a+m+1);
tree->Build_Tree(1,m);
for(i=1;i<=n;i++)
{
switch(operations[i].type)
{
case 1:tree->Get_Mark(1,m,1,0,operations[i].x);break;
case 2:tree->Get_Mark(1,m,1,0,-operations[i].x);break;
case 3:tree->Get_Mark(1,m,operations[i].x,0,0);break;
case 4:tree->Get_Mark(1,m,1,operations[i].x,0);break;
}
if(tree->min_num<L)
tree->Get_L(1,m);
if(tree->max_num>R)
tree->Get_R(1,m);
}
for(i=1;i<=m;i++)
{
int pos=lower_bound(a+1,a+m+1,q[i])-a;
printf("%d\n",tree->Get_Ans(1,m,pos) );
}
return 0;
}
