Points on Cycle （hdu1700，几何）

Points on Cycle

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 1303    Accepted Submission(s): 459

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

 

题意：(摘自网络)一个以原点为中心的圆，告诉你圆上的一个点，求与另外的两个点组成的三角形的周长最长的两点作标。

根据几何知识，知道圆内等边三角形的周长最长。所以题目转化为求已知一个点的圆内接等边三角形的另两点作标。  

思路：设P(x,y),一个方程是x*x+y*y=r*r;另一个方程是根据向量知识，向量的夹角公式得到方程。

因为圆心角夹角为120度，已知一个向量（即一个点作标），所以夹角公式：COS(2PI/3)=a*b/|a|*|b|;(a,b为向量)；

已知角和a向量，就可求b向量b(x,y).由方程组可求得（x,y）；最后得到的是一元二次方程组，可得到两个解，即为两个点的作标。

　　下面是我自己的所作的详解：

　　　　　　　　　　　　　　　　
 

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

#include<stdio.h>
#include<math.h>
int main()
{
    double x1,x2,x3,y1,y2,y3,a,b,c,r;
    int T;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%lf%lf",&x1,&y1);
        r=sqrt(x1*x1+y1*y1);
        a=;
        b=y1;
        c=r*r/-x1*x1;
        y2=(-b-sqrt(b*b-*a*c))/(*a);
        y3=(-b+sqrt(b*b-*a*c))/(*a);
        if(fabs(x1-)<1e-)//注意这里不能是x1==0
        {
            x2=-sqrt(r*r-y2*y2);
            x3=sqrt(r*r-y3*y3);
        }
        else
        {
            x2=(-r*r/-y1*y2)/x1;
            x3=(-r*r/-y1*y3)/x1;
        }
        printf("%.3lf %.3lf %.3lf %.3lf\n",x2,y2,x3,y3);

    }
    return ;
}

我自己错了好几次的，，，，囧rz    写的这么认真，这么详细了，真的很辛苦，

代码如下：