Morley's Theorem UVA

向量函数板
题意：给你一个三角形的三个点abc，求他们的三等分点的交点

#include <iostream>#include <cstring>#include <algorithm>#include <cstdio>#include <cmath>#include <vector>using namespace std;const int N = 13;const int M= 1<<N;int st[M],ma[M];int dp[N][M];struct point{    double x,y;    point(double x=0,double y=0):x(x),y(y){}};typedef point Vector;Vector operator +(point a,point b){    return Vector (a.x+b.x,a.y+b.y);}Vector operator *(point a,double b){    return Vector(a.x*b,a.y*b);}Vector operator -(point a,point b){    return Vector(a.x-b.x,a.y-b.y);}double dot(Vector a,Vector b)//点乘{    return a.x*b.x+a.y*b.y;}double cross(Vector a,Vector b)//叉乘{    return a.x*b.y-a.y*b.x;}double len (Vector a){    return sqrt(a.x*a.x+a.y*a.y);}Vector rotate1(Vector a,double rad)//逆时针旋转rad{    return Vector(a.x*cos(rad)-a.y*sin(rad),a.x*sin(rad)+a.y*cos(rad));}point getans (point p,Vector v,point q,Vector w)//求两条直线的交点，唯一交点，当且仅当不共线{    Vector u=p-q;    double t=cross(w,u)/cross(v,w);    return p+v*t;}point getPoint(point a,point b,point c){    Vector bc=c-b;    Vector ba=a-b;    double rad=acos(dot(bc,ba)/len(bc)/len(ba));    Vector bd=rotate1(bc,rad/3);    Vector cb=b-c;    Vector ca=a-c;    double rad1=acos(dot(cb,ca)/len(cb)/len(ca));    Vector cd=rotate1(cb,-rad1/3);    return getans(b,bd,c,cd);}int main(){    int T;    scanf("%d",&T);    while(T--)    {        point a,b,c;        scanf("%lf %lf %lf %lf %lf %lf",&a.x,&a.y,&b.x,&b.y,&c.x,&c.y);        point d=getPoint(a,b,c);//后两个的 顺序不能写反了，因为有顺时针逆时针的方向问题        point e=getPoint(b,c,a);        point f=getPoint(c,a,b);        printf("%.6f %.6f ",d.x,d.y);        printf("%.6f %.6f ",e.x,e.y);        printf("%.6f %.6f\n",f.x,f.y);    }    return 0;}