HDU 1700 Points on Cycle (坐标旋转)

题目链接：HDU 1700

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

1
2
3

2
1.500 2.000
563.585 1.251

Sample Output

1 2	0.982 -2.299 -2.482 0.299 -280.709 -488.704 -282.876 487.453

Source

2007省赛集训队练习赛（1）

Solution

题意

给定一个圆形在原点的圆和一个圆上的点，在圆上另外找两个点使得三个点组成的三角形周长最大。

思路

坐标旋转

坐标旋转模板题。

三角形是等边三角形是周长最大。

让给定的点绕原点旋转 $120\degree$ 和 $240\degree$ 即可。

Code

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef double db;
const db eps = 1e-10;
const db pi = acos(-1.0);
const ll inf = 0x3f3f3f3f3f3f3f3f;
const ll maxn = 1e5 + 10;

inline int dcmp(db x) {
    if(fabs(x) < eps) return 0;
    return x > 0? 1: -1;
}

class Point {
public:
    double x, y;
    Point(double x = 0, double y = 0) : x(x), y(y) {}
    void input() {
        scanf("%lf%lf", &x, &y);
    }
    bool operator<(const Point &a) const {
        return (!dcmp(y - a.y))? dcmp(x - a.x) < 0: y < a.y;
    }
    Point Rotate(double rad) {
        return Point(x * cos(rad) - y * sin(rad), x * sin(rad) + y * cos(rad));
    }
};

Point ans[2];

int main() {
    int T;
    scanf("%d", &T);
    while(T--) {
        Point p;
        p.input();
        double r = 2.0 * pi / 3.0;
        ans[0] = p.Rotate(r);
        ans[1] = p.Rotate(r * 2);
        sort(ans, ans + 2);
        printf("%.3lf %.3lf %.3lf %.3lf\n", ans[0].x, ans[0].y, ans[1].x, ans[1].y);
    }
    return 0;
}