hdu 4998 矩阵表示旋转

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

http://blog.csdn.net/wcyoot/article/details/33310329

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;}

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