Points on cycle

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

 

 

题目大意：

     给出圆上一点求出该园上任选两点使任意该三角形周长。

 

解题思路：

      最长明显为正三角形周长最长，这题可以用向量旋转公式做，也可以用普通的数学式做。以下给出用向量旋转解题的方法。

 

代码如下：

#include<cmath>
#include<cstdio>
#include<algorithm>
using namespace std;

const double PI = acos(-1.0);
struct Point
{
     double x, y;

     Point(double x = 0, double y = 0): x(x), y(y){}

     void scan()
     {         scanf("%lf%lf", &x, &y);
     }

    void print()
    {
         printf("%.3lf %.3lf", x, y);
     }

     bool operator < (const Point &other){
         return y < other.y || (y == other.y && x < other.x);
     }
 };

 typedef Point Vector;

 Vector rotate(Vector A, double rad)//向量旋转公式
{
     return Vector(A.x * cos(rad) - A.y * sin(rad), A.y * cos(rad) + A.x * sin(rad));
 }

 int main()
 {
     int t;
     Point p[3];
     scanf("%d", &t);
    while(t--)
    {
        p[0].scan();

        p[1] = rotate(p[0], PI * 2 / 3);//逆时针旋转120度
        p[2] = rotate(p[0], -PI * 2 / 3);//顺时针旋转120度

        if(p[2] < p[1]) swap(p[1], p[2]);

         p[1].print(); putchar('  ');
         p[2].print(); putchar('
');
     }
     return 0;
}