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：

 #include <iostream>
 #include <cstdio>
 #include <cstdlib>
 #include <cstring>
 #include <cmath>
 #include <algorithm>
 #include <set>
 #include <map>
 #include <queue>
 #include <vector>
 #include <string>
 #define inf 0x3fffffff
 #define esp 1e-10
 using namespace std;
 void Rotate(double &x0, double &y0, double x, double y, double a)
 {
     double xx, yy;
     xx = x0*cos(a) - y0*sin(a) + x*( - cos(a)) + y*sin(a);
     yy = x0*sin(a) + y0*cos(a) + y*( - cos(a)) - x*sin(a);
     x0 = xx;
     y0 = yy;
 }
 int main()
 {
     //freopen ("test.txt", "r", stdin);
     int T;
     scanf ("%d", &T);
     for (int times = ; times < T; ++times)
     {
         double x0 = , y0 = , x, y, a, p = , ans, pi = 3.14159265358979323846;
         int n;
         scanf ("%d", &n);
         for (int i = ; i < n; ++i)
         {
             scanf ("%lf%lf%lf", &x, &y, &a);
             p += a;
             Rotate(x0, y0, x, y, a);
         }
         while (p > *pi)
         {
             p -= *pi;
         }
         ans = p;
         double k = /sin(p/2.0)/2.0;
         x0 = x0 * k;
         y0 = y0 * k;
         p = pi/2.0 - p/2.0;
         Rotate(x0, y0, , , p);
         printf ("%f %f %f\n", x0, y0, ans);
     }
     return ;
 }