/** * Created by MKD on 2017/8/21. * 思路是不断的利用两个点连线，获取能够取得的最大点，并排除凸点情况 */ import java.util.*; public class Main { public static void main(String[] args){ Scanner in = new Scanner(System.in); int res; int _points_size = 0; _points_size = Integer.parseInt(in.nextLine().trim()); double[] _points = new double[_points_size]; double _points_item; for(int _points_i = 0; _points_i < _points_size; _points_i++) { _points_item = Double.parseDouble(in.nextLine().trim()); _points[_points_i] = _points_item; } res = castle(_points); System.out.println(String.valueOf(res)); } static int castle(double[] points) { int len = points.length; int result = 0; if(len<=8) return 0; else{ for (int j=0;j<len/2 ;j++){ double x1 = points[2*j]; double y1 = points[2*j+1]; for(int k=0;k<len/2;k++){ if(k==j) continue; double x2 = points[2*k]; double y2 = points[2*k+1]; double w= getw(x1,y1,x2,y2); double b = getb(x1,y1,x2,y2); int r= getMax(j,k,len,w,b,points); if(r>result) result=r; } } } return result; } public static Double getw(double x1, double y1, double x2, double y2){//计算w double w = (x1-x2)/(y1-y2); return w; } public static Double getb(double x1, double y1, double x2, double y2){//计算b double w = (x1-x2)/(y1-y2); double b = y1 - x1*w; return b; } public static int getMax(int j,int k,int len,double w,double b,double[] points){//寻找最多的城堡 int n0=0;int n1=0;int n2=0; int result=0; for(int i=0;i<len/2;i++) { if (i == j || i == k) continue; double x = points[2 * i]; double y = points[2 * i + 1]; double re = y - w * x + b; if (re > 0) n1 = n1 + 1; else if (re < 0) n2 = n2 + 1; else n0 = n0 + 1; } result = Math.max((n0+n1),(n2+n0)); if(result == len/2 - 2) return 0;//如果加上连线上除开连接的两个点，数量为total-2，那么一定是凸点 else return result+1;//排除凸点后，总数再加上本身点，固 + 1 } }