Problem Description

There is a cycle with its center on the origin.
Now give you a point on the cycle, you are to find out the other two points on it, to maximize the sum of the distance between each other
you may assume that the radius of the cycle will not exceed 1000.

 

Input

There are T test cases, in each case there are 2 decimal number representing the coordinate of the given point.

 

Output

For each testcase you are supposed to output the coordinates of both of the unknow points by 3 decimal places of precision 
Alway output the lower one first(with a smaller Y-coordinate value), if they have the same Y value output the one with a smaller X.

 

Sample Input

 

Sample Output

 

Source

 

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lcy

 #include <iostream>

 #include <cstdio>

 #include <cmath>

 #include <algorithm>

 using namespace std;

 const double PI = acos(-1.0);

 const double eps = 5e-;

 /****************常用函数***************/

 //判断ta与tb的大小关系

 int sgn( double ta, double tb)

 {

     if(fabs(ta-tb)<eps)return ;

     if(ta<tb)   return -;

     return ;

 }

 //点

 class Point

 {

 public:

     double x, y;

     Point(){}

     Point( double tx, double ty){ x = tx, y = ty;}

 };

 //两点间距离

 double getdis(const Point &st,const Point &se)

 {

     return sqrt((st.x - se.x) * (st.x - se.x) + (st.y - se.y) * (st.y - se.y));

 }

 int main(void)

 {

     //freopen("in.acm","r",stdin);

     int t;

     scanf("%d",&t);

     Point pa,pb,pc,pp=Point(,);

     while(t--)

     {

         scanf("%lf%lf",&pa.x,&pa.y);

         double r = getdis(pa,pp);

         double ag = atan2(pa.y,pa.x);

         pb.x = r * cos(ag + PI *  / ), pb.y = r * sin(ag + PI *  / );

         pc.x = r * cos(ag - PI *  / ), pc.y = r * sin(ag - PI *  / );

         if(sgn(pb.y,pc.y)>||(sgn(pb.y,pc.y)==&&sgn(pb.x,pc.x)>))

             swap(pb,pc);

         printf("%.3f %.3f %.3f %.3f\n",pb.x,pb.y,pc.x,pc.y);

     }

     return ;

 }