题目大意：给定两个三角形，其中一个可以通过以某点为中心旋转并放缩的方式得到另一个，求这个中心それはとっても嬉しいなって。首先两个复数相乘的几何意义是【极角相加，长度相乘】 这两种变换正好对应旋转和放缩那么我们不妨将所有点都放到复平面上由于没有给定点的对应关系，，故我们3!枚举这个对应关系设其中一个三角形的三个顶点为A,B,C，另一个三角形中对应顶点为A’,B’,C’设中心点为P，变换复数为T那么我们有方程组：(A-P)T=(A’-P)(B-P)T=(B’-P)(C-P)T=(C’-P)由前两个方程解得T=(A’-B’)/(A-B)
P=(AT-A’)/(T-1)
代入第三个方程中验证即可。
#include <cmath>#include <cstdio>#include <cstring>#include <iostream>#include <algorithm>#define EPS 1e-4using namespace std;struct Complex{double a,b;Complex() {}Complex(double _,double __=0):a(_),b(__) {}friend istream& operator >> (istream &_,Complex &c){scanf("%lf%lf",&c.a,&c.b);return _;}friend ostream& operator << (ostream &_,const Complex &c){printf("%.6lf %.6lf\n",c.a,c.b);return _;}friend Complex operator + (const Complex &c1,const Complex &c2){return Complex(c1.a+c2.a,c1.b+c2.b);}friend Complex operator – (const Complex &c1,const Complex &c2){return Complex(c1.a-c2.a,c1.b-c2.b);}friend Complex operator * (const Complex &c1,const Complex &c2){return Complex(c1.a*c2.a-c1.b*c2.b,c1.a*c2.b+c1.b*c2.a);}friend Complex operator / (const Complex &c1,const Complex &c2){double modulus=c2.a*c2.a+c2.b*c2.b;return Complex((c1.a*c2.a+c1.b*c2.b)/modulus,(c1.b*c2.a-c1.a*c2.b)/modulus);}friend bool operator == (const Complex &c1,const Complex &c2){return fabs(c1.a-c2.a)<EPS && fabs(c1.b-c2.b)<EPS ;}}A,B,C,points[4];void Calculate(const Complex _A,const Complex _B,const Complex _C){Complex T=(_A-_B)/(A-B),P=(A*T-_A)/(T-1);if( (C-P)*T==(_C-P) ){cout<<P<<endl;throw true;}}int main(){int T;for(cin>>T;T;T–){cin>>A>>B>>C>>points[1]>>points[2]>>points[3];try{Calculate(points[1],points[2],points[3]);Calculate(points[1],points[3],points[2]);Calculate(points[2],points[1],points[3]);Calculate(points[2],points[3],points[1]);Calculate(points[3],points[1],points[2]);Calculate(points[3],points[2],points[1]);}catch(bool){}}}
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