地址：http://acm.split.hdu.edu.cn/showproblem.php?pid=1700

题目：

Points on Cycle

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2523    Accepted Submission(s): 972

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  

 

Source 2007省赛集训队练习赛（1）  

 

Recommend lcy

 思路：

　　直接猜是等边三角形，然后发现确实是。

　　求其他两个点，直接旋转就行了。

#include <iostream>
#include <cstdio>
#include <cmath>
#include <algorithm>

using namespace std;
const double PI = acos(-1.0);
const double eps = 5e-4;

/****************常用函数***************/
//判断ta与tb的大小关系
int sgn( double ta, double tb)
{
if(fabs(ta-tb)<eps)return 0;
if(ta<tb)   return -1;
return 1;
}

//点
class Point
{
public:

double x, y;

Point(){}
Point( double tx, double ty){ x = tx, y = ty;}

};
//两点间距离
double getdis(const Point &st,const Point &se)
{
return sqrt((st.x - se.x) * (st.x - se.x) + (st.y - se.y) * (st.y - se.y));
}

int main(void)
{
//freopen("in.acm","r",stdin);
int t;
scanf("%d",&t);
Point pa,pb,pc,pp=Point(0,0);
while(t--)
{
scanf("%lf%lf",&pa.x,&pa.y);
double r = getdis(pa,pp);
double ag = atan2(pa.y,pa.x);
pb.x = r * cos(ag + PI * 2 / 3), pb.y = r * sin(ag + PI * 2 / 3);
pc.x = r * cos(ag - PI * 2 / 3), pc.y = r * sin(ag - PI * 2 / 3);
if(sgn(pb.y,pc.y)>0||(sgn(pb.y,pc.y)==0&&sgn(pb.x,pc.x)>0))
swap(pb,pc);
printf("%.3f %.3f %.3f %.3f\n",pb.x,pb.y,pc.x,pc.y);
}
return 0;
}