ACM学习历程—Rotate（HDU 2014 Anshan网赛）（几何）

Problem Description

Noting is more interesting than rotation!

Your little sister likes to rotate things. To put it easier to analyze, your sister makes n rotations. In the i-th time, she makes everything in the plane rotate counter-clockwisely around a point ai by a radian of pi.

Now she promises that the total effect of her rotations is a single rotation around a point A by radian P (this means the sum of pi is not a multiplier of 2π).

Of course, you should be able to figure out what is A and P :).

Input

The first line contains an integer T, denoting the number of the test cases.

For each test case, the first line contains an integer n denoting the number of the rotations. Then n lines follows, each containing 3 real numbers x, y and p, which means rotating around point (x, y) counter-clockwisely by a radian of p.

We promise that the sum of all p's is differed at least 0.1 from the nearest multiplier of 2π.

T<=100. 1<=n<=10. 0<=x, y<=100. 0<=p<=2π.

Output

For each test case, print 3 real numbers x, y, p, indicating that the overall rotation is around (x, y) counter-clockwisely by a radian of p. Note that you should print p where 0<=p<2π.

Your answer will be considered correct if and only if for x, y and p, the absolute error is no larger than 1e-5.

Sample Input

1
3
0 0 1
1 1 1
2 2 1

Sample Output

1.8088715944 0.1911284056 3.0000000000






如 图，如果整个屏幕按照A点逆时针旋转a度，就会如图虚线的坐标。根据旋转，可以得到任意一个点旋转后的坐标。还有一点就是旋转最后的综合角度就是直接旋转 角度之和。然后最后得到的O'点，可以通过综合角度a，算出综合旋转中心A。这个旋转中心可以通过先将O‘沿OO’方向移位，使得，OA=OO‘，然后将 O’按照O逆时针旋转到A，求得，旋转角度是pi/2-a/2。

 

代码：

 1 #include <iostream>

 2 #include <cstdio>

 3 #include <cstdlib>

 4 #include <cstring>

 5 #include <cmath>

 6 #include <algorithm>

 7 #include <set>

 8 #include <map>

 9 #include <queue>

10 #include <vector>

11 #include <string>

12 #define inf 0x3fffffff

13 #define esp 1e-10

14 using namespace std;

15 void Rotate(double &x0, double &y0, double x, double y, double a)

16 {

17     double xx, yy;

18     xx = x0*cos(a) - y0*sin(a) + x*(1 - cos(a)) + y*sin(a);

19     yy = x0*sin(a) + y0*cos(a) + y*(1 - cos(a)) - x*sin(a);

20     x0 = xx;

21     y0 = yy;

22 }

23 int main()

24 {

25     //freopen ("test.txt", "r", stdin);

26     int T;

27     scanf ("%d", &T);

28     for (int times = 0; times < T; ++times)

29     {

30         double x0 = 0, y0 = 0, x, y, a, p = 0, ans, pi = 3.14159265358979323846;

31         int n;

32         scanf ("%d", &n);

33         for (int i = 0; i < n; ++i)

34         {

35             scanf ("%lf%lf%lf", &x, &y, &a);

36             p += a;

37             Rotate(x0, y0, x, y, a);

38         }

39         while (p > 2*pi)

40         {

41             p -= 2*pi;

42         }

43         ans = p;

44         double k = 1/sin(p/2.0)/2.0;

45         x0 = x0 * k;

46         y0 = y0 * k;

47         p = pi/2.0 - p/2.0;

48         Rotate(x0, y0, 0, 0, p);

49         printf ("%f %f %f\n", x0, y0, ans);

50     }

51     return 0;

52 }

View Code