# Retis distributed lock (optimistic lock) Nicolos based on LUA script

2022-12-31   ES

## Question description

teleportation door

How to judge?
For every target
ax2+bxy2<=0
ax2+bxy1>=0
x,y1,y2 (For details, please refer to high school mathematics textbook compulsory 5 linear planning issues)
All semi -flat planes are explained and explained
But the precision of this question is very stuck. It is recommended that EPS open a smaller and then use Long Double, and you need to choose two numbers you like when the analytical previous generation value finds the value.(Use the girl’s birthday with caution)

## code

#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
using namespace std;
#define N 200005

const long double eps=1e-12;
const long double inf=1e11;
int dcmp(long double x)
{
if (x<=eps&&x>=-eps) return 0;
return (x>0)?1:-1;
}
struct Vector
{
long double x,y;
Vector(long double X=0,long double Y=0)
{
x=X,y=Y;
}
};
typedef Vector Point;
struct Line
{
Point p;
Vector v;
long double ang;
Line(Point P=Point(0,0),Vector V=Vector(0,0))
{
p=P,v=V;
ang=atan2(v.y,v.x);
}
bool operator < (const Line &a) const
{
return ang<a.ang;
}
};
Vector operator + (Vector a,Vector b) {
return Vector(a.x+b.x,a.y+b.y);}
Vector operator - (Vector a,Vector b) {
return Vector(a.x-b.x,a.y-b.y);}
Vector operator * (Vector a,long double b) {
return Vector(a.x*b,a.y*b);}

int m,n,l,r,ans;
long double x,y,z;
Point P,Q,p[N];
Line s[N],L[N],q[N];

long double Cross(Vector a,Vector b)
{
return a.x*b.y-a.y*b.x;
}
bool Onleft(Line l,Point P)
{
return dcmp(Cross(l.v,P-l.p))>=0;
}
Point GLI(Point P,Vector v,Point Q,Vector w)
{
Vector u=P-Q;
long double t=Cross(w,u)/Cross(v,w);
return P+v*t;
}
bool check(int mid)
{
for (int i=1;i<=mid;++i) L[i]=s[i];
sort(L+1,L+mid+1);
q[l=r=1]=L[1];
for (int i=2;i<=mid;++i)
{
while (l<r&&!Onleft(L[i],p[r-1]))
--r;
while (l<r&&!Onleft(L[i],p[l]))
++l;
q[++r]=L[i];
if (dcmp(Cross(q[r].v,q[r-1].v))==0)
{
--r;
if (Onleft(q[r],L[i].p))
q[r]=L[i];
}
if (l<r)
p[r-1]=GLI(q[r-1].p,q[r-1].v,q[r].p,q[r].v);
}
while (l<r&&!Onleft(q[l],p[r-1]))
--r;
if (r-l<=1) return false;
else return true;
}
int find()
{
int l=5,r=n,mid,ans=4;
while (l<=r)
{
mid=(l+r)>>1;
if (check(mid)) ans=mid,l=mid+1;
else r=mid-1;
}
return ans;
}

int ans = 0; char c; bool flag = false;
while ((c = getchar()) == ' ' || c == '\n' || c == '\r');
if (c == '-') flag = true; else ans = c - '0';
while ((c = getchar()) >= '0' && c <= '9') ans = ans * 10 + c - '0';
return ans * (flag ? -1 : 1);
}

int main()
{
long double a=-1.0,b=4.0;
scanf("%d",&m);
s[++n]=Line(Point(0,0),Vector(0,inf));
s[++n]=Line(Point(0,inf),Vector(-inf,0));
s[++n]=Line(Point(-inf,inf),Vector(0,-inf));
s[++n]=Line(Point(-inf,0),Vector(inf,0));
for (int i=1;i<=m;++i)
{
if (x<=0) break;

P=Point(a,(z-x*x*a)/x);
Q=Point(b,(z-x*x*b)/x);
s[++n]=Line(Q,P-Q);

P=Point(a,(y-x*x*a)/x);
Q=Point(b,(y-x*x*b)/x);
s[++n]=Line(P,Q-P);
}
ans=(find()-4)/2;
printf("%d\n",ans);
}

source