直接6重循环就行了吧。。。判三角形相似直接从小到大枚举两向量夹角是否相等就行了。注意去重点跟三点共线就行了。。。

#include<algorithm>

#include<iostream>

#include<cstring>

#include<cstdlib>

#include<fstream>

#include<sstream>

#include<bitset>

#include<vector>

#include<string>

#include<cstdio>

#include<cmath>

#include<stack>

#include<queue>

#include<stack>

#include<map>

#include<set>

#define FF(i, a, b) for(int i=a; i<b; i++)

#define FD(i, a, b) for(int i=a; i>=b; i--)

#define REP(i, n) for(int i=0; i<n; i++)

#define CLR(a, b) memset(a, b, sizeof(a))

#define debug puts("**debug**")

#define LL long long

#define PB push_back

#define eps 1e-10

using namespace std;

struct Point

{

    double x, y;

    Point (double x=0, double y=0):x(x), y(y) {}

};

typedef Point Vector;

Vector operator + (Vector A, Vector B) { return Vector(A.x + B.x, A.y + B.y); }

Vector operator - (Vector A, Vector B) { return Vector(A.x - B.x, A.y - B.y); }

Vector operator * (Vector A, double p) { return Vector(A.x*p, A.y*p); }

Vector operator / (Vector A, double p) { return Vector(A.x/p, A.y/p); }

bool operator < (const Point& a, const Point& b)

{

    return a.x < b.x || (a.x == b.x && a.y < b.y);

}

int dcmp(double x)

{

    if(fabs(x) < eps) return 0;

    return x < 0 ? -1 : 1;

}

bool operator == (const Point& a, const Point& b)

{

    return dcmp(a.x-b.x) == 0 && dcmp(a.y-b.y) == 0;

}

double Dot(Vector A, Vector B) { return A.x*B.x + A.y*B.y; }

double Length(Vector A) { return sqrt(Dot(A, A)); }

double Angel(Vector A, Vector B) { return acos(Dot(A, B) / Length(A) / Length(B)); }

double Cross(Vector A, Vector B) { return A.x*B.y - A.y*B.x; }

double Area2(Vector A, Vector B, Vector C) { return Cross(B-A, C-A); }

//向量逆时针旋转

Vector Rotate(Vector A, double rad)

{

    return Vector(A.x*cos(rad)-A.y*sin(rad), A.x*sin(rad)+A.y*cos(rad));

}

//求直线p+tv和q+tw的交点 Cross(v, w) == 0无交点

Point GetLineIntersection(Point p, Vector v, Point q, Vector w)

{

    Vector u = p-q;

    double t = Cross(w, u) / Cross(v, w);

    return p + v*t;

}

//点p到直线ab的距离

double DistanceToLine(Point p, Point a, Point b)

{

    Vector v1 = b - a, v2 = p - a;

    return fabs(Cross(v1, v2)) / Length(v1);//如果不带fabs 得到的是有向距离

}

//点p到线段ab的距离

double DistanceToSegment(Point p, Point a, Point b)

{

    if(a == b) return Length(p-a);

    Vector v1 = b-a, v2 = p-a, v3 = p-b;

    if(dcmp(Dot(v1, v2) < 0)) return Length(v2);

    else if(dcmp(Dot(v1, v3)) > 0) return Length(v3);

    else return fabs(Cross(v1, v2)) / Length(v1);

}

//点p在直线ab上的投影

Point GetLineProjection(Point p, Point a, Point b)

{

    Vector v = b-a;

    return a + v*(Dot(v, p-a) / Dot(v, v));

}

//点段相交判定

bool SegmentItersection(Point a1, Point a2, Point b1, Point b2)

{

    double c1 = Cross(a2-a1, b1-a1), c2 = Cross(a2-a1, b2-a1),

    c3 = Cross(b2-b1, a1-b1), c4 = Cross(b2-b1, a2-b1);

    return dcmp(c1)*dcmp(c2) < 0 && dcmp(c3)*dcmp(c4) < 0;

}

//点在线段上

bool OnSegment(Point p, Point a1, Point a2)

{

    return dcmp(Cross(a1-p, a2-p)) == 0 && dcmp(Dot(a1-p, a2-p)) < 0;

}

//多变形面积

double PolygonArea(Point* p, int n)

{

    double ret = 0;

    FF(i, 1, n) ret += Cross(p[i]-p[0], p[i+1]-p[0]);

    return ret/2;

}

Point read_point()

{

    Point a;

    scanf("%lf%lf", &a.x, &a.y);

    return a;

}

bool cmp(double a[], double b[])

{

    if(dcmp(a[0]-b[0])!=0) return 0;

    if(dcmp(a[1]-b[1])!=0) return 0;

    if(dcmp(a[2]-b[2])!=0) return 0;

    return 1;

}

const int maxn = 100;

Point p[maxn];

int n, cnt;

double aa[10], bb[10];

bool can(int i, int j, int k)

{

    return dcmp(Cross(p[j]-p[i], p[k]-p[i])) != 0;

}

int main()

{

    Vector a = Vector(0, 1), b = Vector(0, 100);

    while(scanf("%d", &cnt), cnt)

    {

        REP(i, cnt) p[i] = read_point();

        sort(p, p+cnt);

        n = unique(p, p+cnt) - p;

        int ans = 0;

        REP(i, n) FF(j, i+1, n) FF(k, j+1, n)

        {

             if(!can(i, j, k)) continue;

             int tmp = 0;

             aa[0] = Angel(p[j]-p[i], p[k]-p[i]);

             aa[1] = Angel(p[i]-p[j], p[k]-p[j]);

             aa[2] = Angel(p[i]-p[k], p[j]-p[k]);

             sort(aa, aa+3);

             REP(ii, n) FF(jj, ii+1, n) FF(kk, jj+1, n)

             {

                 if(!can(i, j, k)) continue;

                 bb[0] = Angel(p[jj]-p[ii], p[kk]-p[ii]);

                 bb[1] = Angel(p[ii]-p[jj], p[kk]-p[jj]);

                 bb[2] = Angel(p[ii]-p[kk], p[jj]-p[kk]);

                 sort(bb, bb+3);

                 if(cmp(aa, bb)) tmp++;

             }

             ans = max(ans, tmp);

        }

        printf("%d\n", ans);

    }

    return 0;

}