题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=4998
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
Source
2014 ACM/ICPC Asia Regional Anshan Online
解析系转载：
平面上有一个二维坐标轴上,进行n次操作，把坐标轴绕着(x,y) (这个坐标总是初始坐标轴的坐标) 逆时针转p弧度。
最后的结果相当于进行一次操作，即绕着(X, Y) 逆时针旋转了P弧度。求 X,Y,P，题目保证总有解.
其实可以发现，最后的P是n次的pi的和，因为这和绕什么点旋转无关。（画一个图理解一下吧）
对任意点(x,y)，绕一个坐标点(rx0,ry0)逆时针旋转a角度后的新的坐标设为(x0, y0)，有公式：
x0= (x - rx0)*cos(a) - (y - ry0)*sin(a) + rx0 ;
y0= (x - rx0)*sin(a) + (y - ry0)*cos(a) + ry0 ;
可以假设两个点a, b(两个足够了), 任意指定其坐标, 可以算出旋转后的对应点a‘, b‘,
前面已经求出了P， 现在只要求出旋转中心A即可， 利用 dis(A, a) == dis(A, a‘) 和 dis(A, b) == dis(A, b‘) 直接列方程可以求出A的坐标！
代码如下：
#include <cstdio>
#include <cmath>
#define pi acos(-1.0)
int main()
{
int t;
int n;
double x1, y1, x2, y2, r;
double tx1, ty1, tx2, ty2;
double rx0, ry0, a;
scanf("%d", &t);
while(t--)
{
scanf("%d", &n);
x1 = y1 = 0;
x2 = y2 = 1;
r = 0;
for(int i = 1; i <= n; i++)
{
scanf("%lf%lf%lf", &rx0, &ry0, &a);
r += a;
if(r >= 2*pi)
r -= 2*pi;
tx1 = (x1 - rx0) * cos(a) - (y1 - ry0) * sin(a) + rx0;
ty1 = (x1 - rx0) * sin(a) + (y1 - ry0) * cos(a) + ry0;
tx2 = (x2 - rx0) * cos(a) - (y2 - ry0) * sin(a) + rx0;
ty2 = (x2 - rx0) * sin(a) + (y2 - ry0) * cos(a) + ry0;
x1 = tx1, y1 = ty1;
x2 = tx2, y2 = ty2;
}
double t1 = (x1*x1+y1*y1)*(2*y2-2)-2*y1*(x2*x2+y2*y2)+4*y1;
double t2 = 4*y1-4*y1*x2+4*x1*y2-4*x1;
double x = t1/t2;
double t3 = (x1*x1+y1*y1)-2*x*x1;
double t4 = 2*y1;
double y = t3/t4;
printf("%lf %lf %lf\n",x,y,r);
}
return 0;
}
PS:最后给出求解的方程：
我们设a的坐标为(0, 0)，b的坐标为(1, 1)；
A的坐标为(x, y)；
①：dis(A, a) == dis(A, a‘) ：(x - x1)2 + (y - y1)2 = (x
- 0)2 + (y - 0)2;
②：dis(A, b) == dis(A, b‘) ：(x
- x2)2 + (y - y2)2 = (x - 1)2 + (y - 1)2;
联立①②方程解出A(x,
y)的坐标即可！