题意：

n次旋转  每次平面绕ai点旋转pi弧度  问  最后状态相当于初始状态绕A点旋转P弧度  A和P是多少

思路：

如果初始X点的最后状态为X‘点  则圆心一定在X和X'连线的垂直平分线上  那么仅仅要用在取一个点Y和Y'  相同做它的垂直平分线  两线交点即是圆心  然后用简单几何方法算出角度  最后注意要求最后状态由最初状态逆时针旋转得到  适当调整角度就可以

PS：

kuangbin巨巨的几何代码还是非常easy理解的  非常好用~  谢谢~

代码：

#include<cstdio>

#include<iostream>

#include<string>

#include<cstring>

#include<algorithm>

#include<cmath>

#include<map>

#include<set>

#include<vector>

#include<cassert>

using namespace std;

typedef long long LL;

#define Q 401

#define N 51

#define M 200001

#define inf 2147483647

#define lowbit(x) (x&(-x))

const double eps = 1e-5;

const double PI = acos(-1.0);

int sgn(double x) {

	if (fabs(x) < eps)

		return 0;

	if (x < 0)

		return -1;

	else

		return 1;

}

struct Point {

	double x, y;

	Point() {

	}

	Point(double _x, double _y) {

		x = _x;

		y = _y;

	}

	Point operator +(const Point &b) const {

		return Point(x + b.x, y + b.y);

	}

	Point operator -(const Point &b) const {

		return Point(x - b.x, y - b.y);

	}

	//叉积

	double operator ^(const Point &b) const {

		return x * b.y - y * b.x;

	}

	//点积

	double operator *(const Point &b) const {

		return x * b.x + y * b.y;

	}

	//绕原点逆时针旋转角度B（弧度值）。后x,y的变化

	void transXY(double B) {

		double tx = x, ty = y;

		x = tx * cos(B) - ty * sin(B);

		y = tx * sin(B) + ty * cos(B);

	}

} a1, a2, b1, b2, f1, f2, f3;

struct Line {

	Point s, e;

	Line() {

	}

	Line(Point _s, Point _e) {

		s = _s;

		e = _e;

	}

	//两直线相交求交点

	//第一个值为0表示直线重合，为1表示平行，为2是相交

	//仅仅有第一个值为2时。交点才有意义

	pair<int, Point> operator &(const Line &b) const {

		Point res = s;

		if (sgn((s - e) ^ (b.s - b.e)) == 0) {

			if (sgn((s - b.e) ^ (b.s - b.e)) == 0)

				return make_pair(0, res); //重合

			else

				return make_pair(1, res); //平行

		}

		double t = ((s - b.s) ^ (b.s - b.e)) / ((s - e) ^ (b.s - b.e));

		res.x += (e.x - s.x) * t;

		res.y += (e.y - s.y) * t;

		return make_pair(2, res);

	}

} l1, l2;

//两点间距离

double dist(Point a, Point b) {

	return sqrt((a - b) * (a - b));

}

int t, n;

int main() {

	int i;

	double x, y, z;

	scanf("%d", &t);

	while (t--) {

		a1 = f1 = Point(5e5, 3e5);

		a2 = f2 = Point(1e5, 6e5);

		scanf("%d", &n);

		for (i = 1; i <= n; i++) {

			scanf("%lf%lf%lf", &x, &y, &z);

			f3 = Point(x, y);

			f1 = f1 - f3;

			f2 = f2 - f3;

			f1.transXY(z);

			f2.transXY(z);

			f1 = f1 + f3;

			f2 = f2 + f3;

		}

		b1 = a1 + f1;

		b1.x /= 2;

		b1.y /= 2;

		b2 = f1;

		b2 = b2 - b1;

		b2.transXY(PI / 2);

		b2 = b2 + b1;

		l1 = Line(b1, b2);

		b1 = a2 + f2;

		b1.x /= 2;

		b1.y /= 2;

		b2 = f2;

		b2 = b2 - b1;

		b2.transXY(PI / 2);

		b2 = b2 + b1;

		l2 = Line(b1, b2);

		pair<int, Point> res = l1 & l2;

		i = res.first;

		f3 = res.second;

		assert(i == 2);

		printf("%f %f ", f3.x, f3.y);

		f2 = f1 + a1;

		f2.x /= 2;

		f2.y /= 2;

		x = atan(dist(f1, f2) / dist(f2, f3)) * 2;

		y = PI + PI;

		while (x > y)

			x -= y;

		while (x < 0)

			x += y;

		a1 = a1 - f3;

		a1.transXY(x);

		a1 = a1 + f3;

		a1 = a1 - f1;

		if (sgn(a1.x) || sgn(a1.y))

			x = PI + PI - x;

		printf("%f\n", x);

	}

	return 0;

}