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*(1 - cos(a)) + y*sin(a);
    yy = x0*sin(a) + y0*cos(a) + y*(1 - cos(a)) - x*sin(a);
    x0 = xx;
    y0 = yy;
}
int main()
{
    //freopen ("test.txt", "r", stdin);
    int T;
    scanf ("%d", &T);
    for (int times = 0; times < T; ++times)
    {
        double x0 = 0, y0 = 0, x, y, a, p = 0, ans, pi = 3.14159265358979323846;
        int n;
        scanf ("%d", &n);
        for (int i = 0; i < n; ++i)
        {
            scanf ("%lf%lf%lf", &x, &y, &a);
            p += a;
            Rotate(x0, y0, x, y, a);
        }
        while (p > 2*pi)
        {
            p -= 2*pi;
        }
        ans = p;
        double k = 1/sin(p/2.0)/2.0;
        x0 = x0 * k;
        y0 = y0 * k;
        p = pi/2.0 - p/2.0;
        Rotate(x0, y0, 0, 0, p);
        printf ("%f %f %f\n", x0, y0, ans);
    }
    return 0;
}