HDU 4998 Rotate（鞍山网络赛B题）

HDU 4998 Rotate

题目链接

一个旋转变换可以转化为一个矩阵变化，那么n次对应就是10次矩阵变化，把变化完的矩阵求出来，再逆回去求解答案即可

详细可以看这个博客：二维图形的几何变换

代码：

#include <cstdio>
#include <cstring>
#include <cmath>

const double eps = 1e-8;
const double PI = acos(-1.0);

struct Tran {
	double x, y, r;
	double v[3][3];
	Tran() {
		for (int i = 0; i < 3; i++)
			for (int j = 0; j < 3; j++)
				v[i][j] = 0;
	}
	void init() {
		scanf("%lf%lf%lf", &x, &y, &r);
		for (int i = 0; i < 3; i++)
			for (int j = 0; j < 3; j++)
				v[i][j] = 0;
		v[0][0] = cos(r);
		v[0][1] = sin(r);
		v[1][0] = -sin(r);
		v[1][1] = cos(r);
		v[2][0] = x * (1 - cos(r)) + y * sin(r);
		v[2][1] = y * (1 - cos(r)) - x * sin(r);
		v[2][2] = 1;
	}

	Tran operator * (Tran c) {
		Tran ans;
		for (int i = 0; i < 3; i++)
			for (int j = 0; j < 3; j++)
				for (int k = 0; k < 3; k++)
					ans.v[i][j] += v[i][k] * c.v[k][j];
		return ans;
	}
} tran[15];

int t, n;

int main() {
	scanf("%d", &t);
	while (t--) {
		scanf("%d", &n);
		Tran ans;
		for (int i = 0; i < 3; i++)
			ans.v[i][i] = 1;
		for (int i = 0; i < n; i++) {
			tran[i].init();
			ans = ans * tran[i];
		}
		double r = atan2(ans.v[0][1], ans.v[0][0]);
		if (r < 0) r = 2 * PI + r;
		double cosr = ans.v[0][0];
		double sinr = ans.v[0][1];
		double a1 = 1 - cosr;
		double b1 = sinr;
		double b2 = 1 - cosr;
		double a2 = -sinr;
		double c1 = ans.v[2][0];
		double c2 = ans.v[2][1];
		double y = (c1 * a2 - a1 * c2) / (b1 * a2 - a1 * b2);
		double x = (c1 * b2 - c2 * b1) / (a1 * b2 - a2 * b1);
		printf("%.10lf %.10lf %.10lf\n", x, y, r);
	}
	return 0;
}