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好好领悟一下vector吧!加一行Left还可以判断是否在直线上...

struct L{
D O;V v;db a;
L(){}
L(D O,V v):O(O),v(v){a=atan2(O.y,O.x);}//atan2(y,x)=atan(y/x) but the x is OK to be zero
bl op<(const L&x)const{return a<x.a;}
}; bl Left(D A,L l){return Cross(l.v,A-l.O)>;}
int main(){
V A(,),B(,);
L l(A,B-A);
printf("l( O(%lf,%lf) V(%lf,%lf) a(%lf))\n",l.O.x,l.O.y,l.v.x,l.v.y,l.a);
printf("( 1, 3) is on the %s\n",Left(V( , ),l)?"Left":"right");
printf("( 3, 1) is on the %s\n",Left(V( , ),l)?"Left":"right");
printf("(-1, 3) is on the %s\n",Left(V(-, ),l)?"Left":"right");
printf("( 0,-2) is on the %s\n",Left(V( ,-),l)?"Left":"right");
printf("( 0, 0) is on the %s\n",Left(V( , ),l)?"Left":"right");
return ;
}
l( O(1.000000,1.000000) V(1.000000,1.000000) a(0.785398)) //a=pi/4
( , ) is on the Left
( , ) is on the right
(-, ) is on the Left
( ,-) is on the right
( , ) is on the right

Intersect(L a,L b)

Return The Dot Of Two Line

D Intersect(L a,L b){
V u=a.O-b.O;
return a.O+a.v*(Cross(b.v,u)/Cross(a.v,b.v));
}

int main()

int main(){
L a(D(,),D(,)-D(-,));
L b(D(,),D(,)-D(,));
D A=Intersect(a,b);
cout<<"("<<A.x<<","<<A.y<<")"<<endl;
return ;
}
(0,1)

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