ACM——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

解释：这题主要用到了数学的向量夹角公式，所以这题告诉了我们数学好就是王道。。。

#include<cmath>
#include<cstdio>
#include<algorithm>//交换函数

using namespace std;

const double PI = acos(-1.0);//就是pi的值

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);
    }

};

Point rotate(Point A, double rad)
{
    return Point(A.x * cos(rad) - A.y * sin(rad), A.y * cos(rad) + A.x * sin(rad));//旋转向量公式
}

int main()
{
    int i;
    Point p[3];
    scanf("%d", &i);
    while(i--)
    {
        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].y < p[1].y||(p[2].y == p[1].y && p[2].x < p[1].x))
            swap(p[1], p[2]);//交换

        p[1].print(); //
        printf(" ");

        p[2].print(); //
        printf("\n");
    }
    return 0;
}