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hdu 1700 Points on Cycle 水几何
已知圆心(0,0)圆周上的一点,求圆周上另外两点使得三点构成等边三角形。
懒得推公式,直接用模板2圆(r1=dist,r2=sqrt(3)*dist)相交水过
#include<cstdio> #include<iostream> #include<cmath> #include<algorithm> #include<iterator> using namespace std; #define eps 1e-6 typedef long long ll; inline double sqr(double x) { return x*x; } int gcd(int a,int b) { return b==0?a:gcd(b,a%b); } int sig(double x) { if(fabs(x)<eps) return 0; if(x>0) return 1; return -1; } struct point { double x,y; point(){}; point(double a,double b):x(a),y(b){} void input() { scanf("%lf%lf",&x,&y); } friend point operator + (const point &a,const point &b) { return point(a.x+b.x,a.y+b.y); } friend point operator - (const point &a,const point &b) { return point(a.x-b.x,a.y-b.y); } friend bool operator == (const point &a,const point &b) { return sig(a.x-b.x)==0 && sig(a.y-b.y)==0; } friend point operator * (const point &a,const double &b) { return point(a.x*b,a.y*b); } friend point operator * (const double &a,const point &b) { return point(a*b.x,a*b.y); } friend point operator / (const point &a,const double &b) { return point(a.x/b,a.y/b); } double norm() { return sqrt(sqr(x)+sqr(y)); } }a,b; double det(const point &a,const point &b) { return a.x*b.y-a.y*b.x; } double dot(const point &a,const point &b) { return a.x*b.x+a.y*b.y; } double dist(const point &a,const point &b) { return (a-b).norm(); } point rotate_point(const point &p,double A) { double tx=p.x,ty=p.y; return point(tx*cos(A)-ty*sin(A),tx*sin(A)+ty*cos(A)); } point rotate(const point &p,double cost,double sint) { double x=p.x,y=p.y; return point(x*cost-y*sint,x*sint+y*cost); } pair<point,point> crosspoint(point ap,double ar,point bp,double br) { double d=(ap-bp).norm(); double cost=(ar*ar+d*d-br*br)/(2*ar*d); double sint=sqrt(1.0-cost*cost); point v=(bp-ap)/(bp-ap).norm()*ar; return make_pair(ap+rotate(v,cost,-sint),ap+rotate(v,cost,sint)); } int main() { int cas;b.x=0;b.y=0; scanf("%d",&cas); while(cas--) { a.input(); double r=dist(a,b); pair<point,point> ans=crosspoint(b,r,a,sqrt(3.0)*r); double x1=ans.first.x,y1=ans.first.y,x2=ans.second.x,y2=ans.second.y; if(y1<y2||y1==y2&&x1<x2) printf("%.3lf %.3lf %.3lf %.3lf\n",x1,y1,x2,y2); else printf("%.3lf %.3lf %.3lf %.3lf\n",x2,y2,x1,y1); } return 0; }
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