暑假集训(2)第九弹 ----- Points on Cycle（hdu1700)

                                            Points on Cycle

Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u

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

     

问题分析：它是一道几何题，假设以知点a坐标为（x₀,y₀),位置点b，c为（x₁，y₁)，(x₂,y₂).  圆方程为x²+y² = r²；可将圆方程化为x=rcosα，y=rsinα；

又有x₁²+y₁ ²= r² = x₀²+y₀ ²，(a*b)/|a|*|b| = cos120^。

易得 acosα + bsinα = -0.5r

       （acosα）²   = （0.5r + bsinα）²

        r²sinα² + rbsinα + 0.25r² - a² = 0

      得   x₁ = -0.5*b + a*√3 * 0.5  或 x₁ = -0.5*b - a*√3 *0.5（舍去）

      及   y = -0.5*b - a*√3 * 0.5  或 y= 0.5*b + a*√3 * 0.5（舍去）

同理可得x₂,y₂；

也可直接利用cos(α+β） = cosαcosβ - sinαsinβ ，sin（α+β） = sinαcosβ + cosαsinβ求解，

#include <cstdio>
#include <cmath>
int main()
{
  double a,b,x0,y0,x1,y1,x2,y2;
  int t;
  a=sqrt(3.0)/2;
  b=-0.5;
  scanf("%d",&t);
   while(t--)
  {
    scanf("%lf%lf",&x0,&y0);
    x1 = b*x0 - a*y0;
    y1 = b*y0 + a*x0;
    x2 = b*x0 + a*y0;
    y2 = b*y0 - a*x0;
    if(y1<y2 || ((fabs(y1-y2) < 0.005) && x1 < x2))
      printf("%.3lf %.3lf %.3lf %.3lf
",x1,y1,x2,y2);
    else
      printf("%.3lf %.3lf %.3lf %.3lf
",x2,y2,x1,y1);
  }
  return 0;
}