HDU 4998 Rotate
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; }