Walker

  emmm.......随机化。

  好吧，我们不熟。

  考虑随机选取两组数据高斯消元消除结果后带入检验，能有超过1/2正确就输出。

  其实方程就四个，手动解都没问题。

  只是要注意看sin与cos的关系来确定角象限，被这个卡掉了，挑了好久。

  还要注意在合适的情况下\(eps\)越大越好。

Code

#include<bits/stdc++.h>
using namespace std;
namespace STD
{
	#define rr register
	#define scanf ybbb=scanf
	#define x1 a[id1].X1
	#define x2 a[id2].X1
	#define y1 a[id1].Y1
	#define y2 a[id2].Y1
	#define x1_ a[id1].X2
	#define x2_ a[id2].X2
	#define y1_ a[id1].Y2
	#define y2_ a[id2].Y2
	typedef long long ll;
	const int N=1e5+5;
	const double e=1e-4;
	int n,ybbb	;
	struct line{double X1,X2,Y1,Y2;} a[N];
	int read()
	{
		rr int x_read=0,y_read=1;
		rr char c_read=getchar();
		while(c_read<'0'||c_read>'9')
		{
			if(c_read=='-') y_read=-1;
			c_read=getchar();
		}
		while(c_read<='9'&&c_read>='0')
		{
			x_read=(x_read<<3)+(x_read<<1)+(c_read^48);
			c_read=getchar();
		}
		return x_read*y_read;
	}
	bool check(double cos,double sin,double dx,double dy)
	{
		int cnt=0;
		for(rr int i=1;i<=n;i++)
		{
			double x_=cos*a[i].X1-sin*a[i].Y1+dx;
			double y_=cos*a[i].Y1+sin*a[i].X1+dy;
			if(fabs(x_-a[i].X2)<=e&&fabs(y_-a[i].Y2)<=e)
				cnt++;
		}
		return cnt>=((n+1)>>1)&&cnt<=n;
	}
};
using namespace STD;
int main()
{
	n=read();
	for(rr int i=1;i<=n;i++)
		scanf("%lf%lf%lf%lf",&a[i].X1,&a[i].Y1,&a[i].X2,&a[i].Y2);
	double x,b,c,d,scale,cos,sin;
	srand(time(0));
	for(rr int i=1;i<=500;i++)
	{
		int id1=rand()%n+1;
		int id2=rand()%n+1;
		while(id1==id2) id2=(id2-rand()%n+n)%n+1;
		x=x1_-x2_-(y1_-y2_)*(y2-y1)/(x1-x2);
		x=x/(x1-x2+(y1-y2)*(y1-y2)/(x1-x2));
		b=((y1_-y2_)-(x*(y1-y2)))/(x1-x2);
		c=(x1_-x*x1+b*y1);d=(y1_-b*x1-x*y1);
		if(!check(x,b,c,d))continue;
		scale=sqrt(x*x+b*b);
		cos=x/scale,sin=b/scale;
		break;
	}
	printf("%.7lf\n%.7lf\n%.7lf %.7lf\n",sin>0?acos(cos):-acos(cos),scale,c,d);
}