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;
}