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hdu1700 Points on Cycle

时间:2017-10-12 15:45:25      阅读:167      评论:0      收藏:0      [点我收藏+]

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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
 

 

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。

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cmath>
 4 #include <algorithm>
 5 
 6 
 7 using namespace std;
 8 const double PI = acos(-1.0);
 9 const double eps = 5e-4;
10 
11 /****************常用函数***************/
12 //判断ta与tb的大小关系
13 int sgn( double ta, double tb)
14 {
15     if(fabs(ta-tb)<eps)return 0;
16     if(ta<tb)   return -1;
17     return 1;
18 }
19 
20 //
21 class Point
22 {
23 public:
24 
25     double x, y;
26 
27     Point(){}
28     Point( double tx, double ty){ x = tx, y = ty;}
29 
30 };
31 //两点间距离
32 double getdis(const Point &st,const Point &se)
33 {
34     return sqrt((st.x - se.x) * (st.x - se.x) + (st.y - se.y) * (st.y - se.y));
35 }
36 
37 int main(void)
38 {
39     //freopen("in.acm","r",stdin);
40     int t;
41     scanf("%d",&t);
42     Point pa,pb,pc,pp=Point(0,0);
43     while(t--)
44     {
45         scanf("%lf%lf",&pa.x,&pa.y);
46         double r = getdis(pa,pp);
47         double ag = atan2(pa.y,pa.x);
48         pb.x = r * cos(ag + PI * 2 / 3), pb.y = r * sin(ag + PI * 2 / 3);
49         pc.x = r * cos(ag - PI * 2 / 3), pc.y = r * sin(ag - PI * 2 / 3);
50         if(sgn(pb.y,pc.y)>0||(sgn(pb.y,pc.y)==0&&sgn(pb.x,pc.x)>0))
51             swap(pb,pc);
52         printf("%.3f %.3f %.3f %.3f\n",pb.x,pb.y,pc.x,pc.y);
53     }
54     return 0;
55 }

 

hdu1700 Points on Cycle

标签:with   ons   between   freopen   title   pos   scanf   recommend   scan   

原文地址:http://www.cnblogs.com/weeping/p/7656195.html

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