hdu1700 Points on Cycle (数学)

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.

NOTE

when output, if the absolute difference between the coordinate values X1 and X2 is smaller than 0.0005, we assume they are equal.

 

 

Sample Input

2 1.500 2.000 563.585 1.251

 

 

Sample Output

0.982 -2.299 -2.482 0.299 -280.709 -488.704 -282.876 487.453

#include<stdio.h>
#include<math.h>
int main()
{
    int t;
    double x[3],y[3],c[2],a,b;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%lf%lf",&x[0],&y[0]);
        c[0]=x[0]*x[0]+y[0]*y[0];
        if(x[0]==0)
        {
            y[1]=y[2]=-y[0]/2;
            x[1]=-sqrt(c[0]-y[1]*y[1]);
            x[2]=sqrt(c[0]-y[2]*y[2]);
        }
        else
        {
            c[1]=-c[0]+c[0]*c[0]/(4*x[0]*x[0]);
            b=c[0]*y[0]/(x[0]*x[0]);
            a=1+y[0]*y[0]/(x[0]*x[0]);
            y[1]=(-b-sqrt(b*b-4*a*c[1]))/(2*a);
            y[2]=(-b+sqrt(b*b-4*a*c[1]))/(2*a);
            x[1]=-c[0]/(2*x[0])-y[0]*y[1]/x[0];
            x[2]=-c[0]/(2*x[0])-y[0]*y[2]/x[0];
        }
        printf("%.3lf %.3lf %.3lf %.3lf\n",x[1],y[1],x[2],y[2]);
    }
}