题意：给出n,m。
n表示给出的n个横坐标为1-n，y为0的坐标m表示以下有m个坐标，在横坐标上的点
向各个角度看，在可以看到最多的点在同一条直线上的点的做多值为横坐标这一点的值，最后各个
横坐标的值的和为多少
思路：由于m的值为枚举随意的两个点连成的直线，看在直线上的点有多少，看这条线和横坐标的值为
多少。是否是整值点。假设是就记录这个整值点的最大值
*/
#include<bits/stdc++.h>
using namespace std;
const int maxn = 1e6 +5;
const double eps =1e-3;
int b[maxn];
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 +(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);
}
double Cross(Vector A,Vector B)
{
return A.x*B.y-A.y*B.x;
}
int dcmp(double x)
{
if(fabs(x)<eps) return 0;
return x>eps?
1:-1;
}
struct Line
{
Point p;
Vector v;
};
int GetLineIntersection(Line L1,Line L2)
{
if(dcmp(Cross(L1.v,L2.v))==0)
return 0;
Vector u=L1.p-L2.p;
int t=Cross(L2.v,u)/Cross(L1.v,L2.v);
Point answer;
answer=L1.p+L1.v*t;
int temp=(int)(answer.x+0.5);
if(dcmp(answer.x)>0&&dcmp(answer.x-temp)==0&&dcmp(answer.y)==0&&temp<=1000000)
return temp;
return 0;
}
int main()
{
int n,m;
Point a[300];
while(scanf("%d%d",&n,&m)!=EOF)
{
memset(b,0,sizeof(b));
for(int i=0; i<m; i++)
scanf("%lf%lf",&a[i].x,&a[i].y);
Line ox;
ox.p.x=0;
ox.p.y=0;
ox.v.x=1;
ox.v.y=0;
Line l1;
for(int i=0; i<m; i++)
{
for(int j=i+1; j<m; j++)
{
int have=2;
for(int k=j+1; k<m; k++)
{
if(dcmp(Cross(a[i]-a[j],a[k]-a[j]))==0)
have++;
}
l1.p=a[i];
l1.v=a[i]-a[j];
int flag=GetLineIntersection(ox,l1);
if(flag)
{
a[299].x=flag;
a[299].y=0;
if(dcmp(Cross(a[i]-a[j],a[299]-a[j]))==0)
b[flag]=max(b[flag],have);
}
}
}
int sum=0;
for(int i=1; i<=n; i++)
if(b[i])
sum+=b[i];
else
sum++;
cout << sum << endl;
}
return 0;
}