标签：des style blog class code java
Several days ago, a beast caught a beautiful princess and the princess was put in prison. To rescue the princess, a prince who wanted to marry the princess set out immediately. Yet, the beast set a maze. Only if the prince find out the maze’s exit can he save the princess.
Now, here comes the problem. The maze is a dimensional plane. The beast is smart, and he hidden the princess snugly. He marked two coordinates of an equilateral triangle in the maze. The two marked coordinates are A(x1,y1) and B(x2,y2). The third coordinate C(x3,y3) is the maze’s exit. If the prince can find out the exit, he can save the princess. After the prince comes into the maze, he finds out the A(x1,y1) and B(x2,y2), but he doesn’t know where the C(x3,y3) is. The prince need your help. Can you calculate the C(x3,y3) and tell him?
The first line is an integer T(1 <= T <= 100)
which is the number of test cases. T test cases follow. Each test case contains
two coordinates A(x1,y1) and
B(x2,y2), described by four floating-point numbers
x1, y1, x2, y2 (
|x1|, |y1|, |x2|,
|y2| <= 1000.0).
Please notice that
A(x1,y1) and B(x2,y2) and
C(x3,y3) are in an anticlockwise direction from
the equilateral triangle. And coordinates
A(x1,y1) and B(x2,y2) are given by
anticlockwise.
4 -100.00 0.00 0.00 0.00 0.00 0.00 0.00 100.00 0.00 0.00 100.00 100.00 1.00 0.00 1.866 0.50
(-50.00,86.60) (-86.60,50.00) (-36.60,136.60) (1.00,1.00)
1 #include <iostream>
2 #include <stdio.h>
3 #include <cmath>
4 using namespace std;
5 #define eps 1e-10
6 #define PI acos(-1)
7 struct Point{
8 double x,y;
9 Point(double x=0,double y=0):x(x),y(y){}
10 };
11 typedef Point Vector ;
12 Vector operator + (Vector a,Vector b)
13 {
14 return Vector(a.x+b.x,a.y+b.y);
15 }
16 Vector operator - (Point a,Point b)
17 {
18 return Vector(a.x-b.x,a.y-b.y);
19 }
20 Vector operator * (Vector a,double b)
21 {
22 return Vector(a.x*b,a.y*b);
23 }
24 Vector operator / (Vector a,double b)
25 {
26 return Vector(a.x/b,a.y/b);
27 }
28 double Cross(Vector a,Vector b)
29 {
30 return a.x*b.y-b.x*a.y;
31 }
32 Vector Rotate(Vector A,double rad)
33 {
34 return Vector(A.x*cos(rad)-A.y*sin(rad),A.x*sin(rad)+A.y*cos(rad));
35 }
36 Point GetLineIntersection(Point P,Vector v,Point Q,Vector w)
37 {
38 Vector u = P-Q;
39 double t = Cross(w,u) / Cross(v,w);
40 return P+v*t;
41 }
42 int main()
43 {
44 int T;
45 scanf("%d",&T);
46 while(T--){
47 Point a,b;
48 scanf("%lf%lf%lf%lf",&a.x,&a.y,&b.x,&b.y);
49 Point c = GetLineIntersection(a,Rotate(b-a,PI/3),b,Rotate(b-a,PI*2/3));
50 printf("(%.2lf,%.2lf)\n",c.x,c.y);
51 }
52 return 0;
53 }
54
55
56
57
58 /**************************************
59 Problem id : SDUT OJ 2603
60 User name : Miracle
61 Result : Accepted
62 Take Memory : 512K
63 Take Time : 0MS
64 Submit Time : 2014-05-04 09:16:11
65 **************************************/
Freecode : www.cnblogs.com/yym2013
sdut 2603:Rescue The Princess（第四届山东省省赛原题，计算几何，向量旋转 + 向量交点）,布布扣,bubuko.com
sdut 2603:Rescue The Princess（第四届山东省省赛原题，计算几何，向量旋转 + 向量交点）
标签：des style blog class code java
原文地址：http://www.cnblogs.com/yym2013/p/3707785.html
