hdu4998 旋转坐标系

题意：

      一开始的时候有一个坐标系（正常的），然后有n个操作，每个操作是 x y d,意思是当前坐标系围绕x,y点逆时针旋转d度，最后让你输出三个数x y d，把这n个操作的最后结果，用一步等效过来，就是找到一个点，逆时针旋转一个度数，等于当前的这个状态。

思路：

      我们可以用一个向量来代表当前坐标系，每次操作把当前向量拆成两个点单独操作，假如当前向量a,b，绕点c旋转d度，那么我们可以等效向量c,a逆时针旋转d，然后向量c,b逆时针旋转d，这样就的到了两个新的向量，此时我们要根据这两个新的向量求出当前这两个点的新位置，然后再用当前的新位置和下一组操作，最后得到了最终的一个向量，现在我们只要求出初始向量和最终向量的转换关系就行了，这个地方首先我们求转换点，求法是两个向量的x，x'连线，y.y'连线，两条线段中垂线的交点，求出交点之后再用余弦定理求出夹角，然后在用向量的关系来判断要不要用2PI-当前度数，具体看代码。

#include<math.h>
#include<algorithm>
#include<stdio.h>
#define maxn 60
#define eps 1e-7
#define PP (3.141592653589793238)
using namespace std;

int dcmp(double x)
{
    if(fabs(x)<eps) return 0;
    else return x<0?-1:1;
}
double toRad(double deg)
{
    return deg/180.0*acos(-1.0);
}
struct Point
{
    double x,y;
    Point(){}
    Point(double x,double y):x(x),y(y) {}
    void input()
    {
        scanf("%lf %lf",&x,&y);
    }
};
typedef Point Vector;

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, double p )
{
    return Vector( A.x * p, A.y * p );
}
Vector operator/( Vector A, double p )
{
    return Vector( A.x / p, A.y / p );
}
bool operator<(const Point& A, const Point& B )
{
    return dcmp( A.x - B.x ) < 0 || ( dcmp( A.x - B.x ) == 0 && dcmp( A.y - B.y ) < 0 );
}
bool operator==( const Point& a, const Point& b )
{
    return dcmp( a.x - b.x ) == 0 && dcmp( a.y - b.y ) == 0;
}
struct Line
{
    Point s,e;
    Vector v;
    Line() {}
    Line(Point s,Point v,int type):
        s(s),v(v){}
    Line(Point s,Point e):s(s),e(e)
    {v=e-s;}

};
double Dot(Vector A,Vector B)
{
    return A.x*B.x+A.y*B.y;
}
double Length(Vector A)
{
    return sqrt(Dot(A,A));
}
double Angle(Vector A,Vector B)
{
    return acos(Dot(A,B)/Length(A)/Length(B));
}
double Cross(Vector A,Vector B)
{
    return A.x*B.y-A.y*B.x;
}
double Area2(Point A,Point B,Point C )
{
    return Cross(B-A,C-A);
}
double Dist(Point A,Point B)
{
    return Length(A-B);
}
Vector Rotate(Vector A, double rad)
{
    return Vector(A.x*cos(rad)-A.y*sin(rad),A.x*sin(rad)+A.y*cos(rad));
}
Vector Normal(Vector A)
{
    double L=Length(A);
    return Vector(-A.y/L,A.x/L);
}
Point GetLineIntersection(Line l1,Line l2)
{
    Point P=l1.s;
    Vector v=l1.v;
    Point Q=l2.s;
    Vector w=l2.v;
    Vector u=P-Q;
    double t=Cross(w,u)/Cross(v,w);
    return P+v*t;
}
double DistanceToLine(Point P,Line L)
{
    Point A,B;
    A=L.s,B=L.e;
    Vector v1=B-A,v2=P-A;
    return fabs(Cross(v1,v2))/Length(v1);
}
double DistanceToSegment(Point P, Line L)
{
    Point A,B;
    A=L.s,B=L.e;
    if(A==B) return Length(P-A);
    Vector v1=B-A,v2=P-A,v3=P-B;
    if (dcmp(Dot(v1,v2))<0) return Length(v2);
    else if (dcmp(Dot(v1,v3))>0) return Length(v3);
    else return fabs(Cross(v1,v2)) / Length(v1);
}
Point GetLineProjection(Point P,Line L)
{
    Point A,B;
    A=L.s,B=L.e;
    Vector v=B-A;
    return A+v*(Dot(v,P-A)/Dot(v,v));
}

double abss(double x)
{
   return x < 0 ? -x : x;
}

bool OnSegment(Point p,Line l)
{
    Point a1=l.s;
    Point a2=l.e;
    return dcmp(Cross(a1-p,a2-p))==0&&dcmp(Dist(p,a1)+Dist(p,a2)-Dist(a1,a2))==0;
}
bool Paralled(Line l1,Line l2)
{
    return dcmp(Cross(l1.e-l1.s,l2.e-l2.s))==0;
}
bool SegmentProperIntersection(Line l1,Line l2)
{
    if(Paralled(l1,l2))
    {
        return false;
    }
    Point t=GetLineIntersection(l1,l2);
    if(OnSegment(t,l1))
    {
        return true;
    }
    return false;
}

int main ()
{
   double x ,y ,p;
   int T ,n ,i;
   scanf("%d" ,&T);
   while(T--)
   {
      scanf("%d" ,&n);
      double nowx1  = 0 ,nowy1 = 0;
      double nowx2  = 0 ,nowy2 = 101.0;
      double sss = 0;;
      Vector A ,B;
      for(i = 1 ;i <= n ;i ++)
      {
         scanf("%lf %lf %lf" ,&x ,&y ,&p);
         if(p == 0.0 || abss(p - PP * 2) <= 0.00001) continue;
         sss += p;
         A.x = nowx1 - x ,A.y = nowy1 - y;
         B = Rotate(A ,p);
         nowx1 = x + B.x ,nowy1 = y + B.y;
         A.x = nowx2 - x ,A.y = nowy2 - y;
         B = Rotate(A ,p);
         nowx2 = x + B.x ,nowy2 = y + B.y;
      }
      if(nowx1 == 0.0 && nowy1 == 0.0)
      {
          double x4 = nowx2 ,y4 = nowy2;
          double x3 = 0 ,y3 = 0;
          double x1 = 0 ,y1 = 101.0;
          double aaa;
          double tmp = (x4 - x3) * (x1 - x3) + (y4 - y3) * (y1 - y3);
          tmp = tmp / (pow(x4 - x3 ,2.0) + pow(y4 - y3 ,2.0));
          aaa = acos(tmp);
          double q1 = 0 ,q2 = 0;
          if(nowx2 > 0.0) aaa = PP * 2 - aaa;
          if(abss(aaa - PP * 2) <= 0.00001)aaa = 0;
          printf("%lf %lf %lf\n" ,q1 ,q2 ,aaa);
      }
      else if(nowx2 == 0.0 && nowy2 == 101.0)
      {
          double x4 = nowx1 ,y4 = nowy1;
          double x3 = 0 ,y3 = 101.0;
          double x1 = 0 ,y1 = 0;
          double aaa;
          double tmp = (x4 - x3) * (x1 - x3) + (y4 - y3) * (y1 - y3);
          tmp = tmp / (pow(x4 - x3 ,2.0) + pow(y4 - y3 ,2.0));
          aaa = acos(tmp);
          double q1 = 0 ,q2 = 101.0;
          if(nowx1 < 0) aaa = PP * 2 - aaa;
          if(abss(aaa - PP * 2) <= 0.00001)aaa = 0;
          printf("%lf %lf %lf\n" ,q1 ,q2 ,aaa);
      }
      else
      {
         Point AA1;
         AA1.x = AA1.y = 0;
         Point BB1;
         BB1.x = nowx1 ,BB1.y = nowy1;
         Line now1 = Line((AA1 + BB1)/2 ,Normal(AA1 - BB1),1);
         Point AA2;
         AA2.x = 0 ,AA2.y = 101.0;
         Point BB2;
         BB2.x = nowx2 ,BB2.y = nowy2;
         Line now2 = Line((AA2 + BB2)/2 ,Normal(AA2 - BB2),1);
         Point now = GetLineIntersection(now1 ,now2);
         double x4 = nowx1 ,y4 = nowy1;
         double x3 = now.x ,y3 = now.y;
         double x1 = 0 ,y1 = 0;
         double aaa;
         double tmp = (x4 - x3) * (x1 - x3) + (y4 - y3) * (y1 - y3);
         tmp = tmp / (pow(x4 - x3 ,2.0) + pow(y4 - y3 ,2.0));
         double x2 ,y2;
         x1 = 0 ,y1 = 101;
         x2 = nowx2 - nowx1 ,y2 = nowy2 - nowy1;
         aaa = acos(tmp);
         if(x1*y2-x2*y1<0) aaa = PP * 2 - aaa;
         printf("%lf %lf %lf\n" ,now.x ,now.y ,aaa);
     }
   }
   return 0;
}