数学题，看我的程序吧。我的注释打得很全。
还有，就是最后有用到高斯消元来解方程。
CO
/*
PROGRAM: $PROGRAM
AUTHOR: Su Jiao
DATE: 2010-3-26
DESCRIPTION:
$DESCRIPTION
*/
#include <iostream>
using std::cin;
using std::cout;
#include <fstream>
using std::ifstream;
using std::ofstream;
#include <sstream>
using std::stringstream;
using std::endl;
#include <vector>
using std::vector;
#include <string>
using std::string;
#include <stack>
using std::stack;
#include <queue>
using std::queue;
#include <set>
using std::set;
#include <map>
using std::map;
using std::pair;
using std::make_pair;
#include <algorithm>
using std::sort;
#include <cassert>
//using std::assert;
#include <cmath>
using std::cin;
using std::cos;
//using std::M_PI;
#define M_PI 3.141592653589793238462643383279f
#include <iomanip>
template<typename Type>
class Function
{
int n;
vector<vector<Type> > matrix;
vector<Type> answer;
static Type abs(Type a)
{
return a>=0?a:-a;
}
void smaller(int step)
{
for (int i=step+1;i<n;i++)
if (abs(matrix[i][step])>abs(matrix[step][step]))
matrix[step].swap(matrix[i]);
for (int i=step+1;i<n;i++)
{
Type mul=matrix[i][step];
Type div=matrix[step][step];
for (int j=step;j<=n;j++)
matrix[i][j]-=matrix[step][j]*mul/div;
}
}
void back(int step)
{
if (step==n-1)
answer[step]=matrix[step][n]/matrix[step][step];
else
{
answer[step]=matrix[step][n];
for (int i=step+1;i<n;i++)
answer[step]-=matrix[step][i]*answer[i];
answer[step]/=matrix[step][step];
}
}
public:
void read(vector<vector<Type> >& get)
{
n=get.size();
matrix=get;
answer.resize(get.size());
}
void solve()
{
for (int i=0;i<n-1;i++) smaller(i);
for (int i=n-1;i>=0;i–) back(i);
}
void write(vector<Type>& put)
{
put=answer;
}
};
class Application
{
typedef double Type;
struct Da
{
typedef double Type;
Type x,y,C;
void set(Type _x,Type _y,Type _C)
{
x=(_x),y=(_y),C=(_C);
}
void print()
{
cout<<floor(x+0.5)<<” “<<floor(y+0.5)<<” “<<floor(C+0.5)<<endl;
}
};
int N;
vector<pair<Type,Type> > M;
vector<Type> a;
public:
Application()
{
cin>>N;
M.resize(N);
for (int i=0;i<N;i++)
cin>>M[i].first>>M[i].second;
a.resize(N);
for (int i=0;i<N;i++)
{
cin>>a[i];
a[i]=a[i]/180.0*M_PI;
}
}
int run()
{
/*
O为原点 旋转均为顺时针
O->Ai+1= Mi->Ai 转ai度后 + O->Mi
Ai+1=(Ai-Mi)转ai度 + Mi
然后，关于旋转：
若(x,y)旋转a度成为了(x’,y’)
可以用极坐标
令x’=R*cos(A),y’=R*sin(A)
x=R*cos(B),y=R*cos(B)
则A-B=a
所以cos(A)=cos(B+a)
sin(A)=sin(B+a)
那么cos(A)=cos(B)cos(a)-sin(B)sin(a)
sin(A)=sin(B)cos(a)+cos(B)sin(a)
所以x’=x*cos(a)-y*sin(a)
y’=y*cos(a)+x*sin(a)
所以x(Ai+1)=(x(Ai)-x(Mi))*cos(a)-(y(Ai)-y(Mi))*sin(a)+x(Mi)
y(Ai+1)=(y(Ai)-y(Mi))*cos(a)+(x(Ai)-x(Mi))*sin(a)+y(Mi)
所以x(Ai+1)=x(Ai)*cos(a)-y(Ai)*sin(a)-x(Mi)*cos(a)+y(Mi)*sin(a)+x(Mi)
y(Ai+1)=y(Ai)*cos(a)+x(Ai)*sin(a)-y(Mi)*cos(a)-x(Mi)*sin(a)+y(Mi)
*/
vector<pair<Da
A[0].first.set(1.0,0,0);
A[0].second.set(0,1.0,0);
for (int i=0;i<N;i++)
{
double x,y,C;
//x(Ai+1)=x(Ai)*cos(a) -y(Ai)*sin(a)
x=A[i].first.x*cos(a[i])-A[i].second.x*sin(a[i]);
y=A[i].first.y*cos(a[i])-A[i].second.y*sin(a[i]);
C=A[i].first.C*cos(a[i])-A[i].second.C*sin(a[i])
//-x(Mi)*cos(a) +y(Mi)*sin(a) +x(Mi)
-M[i].first*cos(a[i])+M[i].second*sin(a[i])+M[i].first;
A[i+1].first.set(x,y,C);
//y(Ai)*cos(a) +x(Ai)*sin(a)
x=A[i].second.x*cos(a[i])+A[i].first.x*sin(a[i]);
y=A[i].second.y*cos(a[i])+A[i].first.y*sin(a[i]);
C=A[i].second.C*cos(a[i])+A[i].first.C*sin(a[i])
//-y(Mi)*cos(a) -x(Mi)*sin(a) +y(Mi)
-M[i].second*cos(a[i])-M[i].first*sin(a[i])+M[i].second;
A[i+1].second.set(x,y,C);
}
Function<Type> f;
vector<vector<double> > matrix(2,vector<double>(3));
vector<double> answer(2);
//x=A[N].first.x*x+A[N].first.y*y+A[N].first.C
//y=A[N].second.x*x+A[N].second.y*y+A[N].first.C
// x y C
// 1-A[N].first.x -A[N].first.y A[N].first.C
// -A[N].second.x 1-A[N].second.y A[N].second.C
matrix[0][0]=1-A[N].first.x;
matrix[0][1]=-A[N].first.y;
matrix[0][2]=A[N].first.C;
matrix[1][0]=-A[N].second.x;
matrix[1][1]=1-A[N].second.y;
matrix[1][2]=A[N].second.C;
f.read(matrix);
f.solve();
f.write(answer);
double X=answer[0];
double Y=answer[1];
cout<<std::setiosflags(std::ios::fixed)<<std::setprecision(2);
for (int i=0;i<N;i++)
{
cout<<A[i].first.x*X+A[i].first.y*Y+A[i].first.C<<” “
<<A[i].second.x*X+A[i].second.y*Y+A[i].second.C<<endl;
}
return 0;
}
};
int main()
{
Application app;
return app.run();
}