hdu 4998 矩阵表示旋转

http://acm.hdu.edu.cn/showproblem.php?pid=4998

/article/1412788.html

一个旋转变换可以转化为一个三维矩阵的变化

绕(x,y)旋转角度r，执行十次，求等价旋转点和角度

绕原点矩阵如下



由于是绕(x,y)，x1 = (x-x0)*cos0 - (y-y0)*sin0
+ x0;y1同理，那么第三行前两列即为x0*(1-cos(r)) + y0*sin(r)和y0*(1-cos(r)) - x0*sin(r)

最后根据x0*(1-cos(r))
+ y0*sin(r) = v[2][0]和y0*(1-cos(r))
- x0*sin(r) = v[2][1]列出方程即可求解等价的x0,y0

#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <cstring>
#include <string>
#include <queue>
#include <map>
#include <iostream>
#include <algorithm>
using namespace std;
#define RD(x) scanf("%d",&x)
#define RD2(x,y) scanf("%d%d",&x,&y)
#define RD3(x,y,z) scanf("%d%d%d",&x,&y,&z)
#define clr0(x) memset(x,0,sizeof(x))
typedef long long LL;
double x,y,r;
const double pi = acos(-1.0);
struct Matrix
{
double v[3][3];
Matrix(){
for(int i = 0;i < 3;++i)
for(int j = 0;j < 3;++j)
v[i][j] = 0;
}
void id(){
for(int i = 0;i < 3;++i)
v[i][i] = 1;
}
void init(){
v[1][1] = v[0][0] = cos(r);
v[1][0] = -(v[0][1] = sin(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;
}
Matrix operator * (Matrix c){
Matrix 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;
}
};

int main() {
int _,n;RD(_);while(_--){
RD(n);
Matrix ans,tmp[11];
ans.id();
for(int i = 0;i < n;++i){
scanf("%lf%lf%lf", &x, &y, &r);
tmp[i].init();
ans = ans*tmp[i];
}
double cosr = ans.v[0][0],sinr = ans.v[0][1];
r = atan2(sinr,cosr);
if(r < 0)
r += 2*pi;
double c1 = ans.v[2][0],c2 = ans.v[2][1];
double y = (c2*(cosr - 1) - sinr*c1)/(-sinr*sinr-(1-cosr)*(1-cosr)),
x = (c1*(1-cosr) - c2*sinr)/((1-cosr)*(1-cosr) + sinr*sinr);
printf("%.10lf %.10lf %.10lf\n", x, y, r);
}
return 0;
}