因為是x，y均為整數因此對於同一直線的點，其最簡分數x/y是相同的(y可以為0，這裏不做除法)於是將這些點不斷求最簡分數用pair在set中去重即可。

#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cassert>
#include <cstring>
#include <set>
#include <map>
#include <list>
#include <queue>
#include 
 
<string>
#include <iostream>
#include <algorithm>
#include <functional>
#include <stack>
using namespace std;
typedef long long ll;
#define T int t_;Read(t_);while(t_--)
#define dight(chr) (chr>=‘0‘&&chr<=‘9‘)
#define alpha(chr) (chr>=‘a‘&&chr<=‘z‘)
#define
 
 INF (0x3f3f3f3f)
#define maxn (100005)
#define hashmod 100000007
#define ull unsigned long long
#define repne(x,y,i) for(i=x;i<y;++i)
#define repe(x,y,i) for(i=x;i<=y;++i)
#define ri register int
void Read(int &n){char chr=getchar(),sign=1;for(;!dight(chr);chr=getchar())if(chr==‘-‘)sign=-1;
    
 
for(n=0;dight(chr);chr=getchar())n=n*10+chr-‘0‘;n*=sign;}
void Read(ll &n){char chr=getchar(),sign=1;for(;!dight(chr);chr=getchar())if
    (chr==‘-‘)sign=-1;
    for(n=0;dight(chr);chr=getchar())n=n*10+chr-‘0‘;n*=sign;}
int n,x,y,a,b;
int gcd(int x,int y){return y==0?x:gcd(y,x%y);}
set<pair<int,int> > se1,se2;
int main()
{
    freopen("a.in","r",stdin);
    freopen("b.out","w",stdout);
    ri i;
    Read(n),Read(x),Read(y);
    repe(1,n,i){
        Read(a),Read(b);
        int ex = abs(x - a),ey = abs(y - b),g = gcd(ex,ey);
        if((x - a >= 0 && y - b >= 0) || (x - a <= 0 && y - b <= 0)) se1.insert(make_pair(ex/g,ey/g));
        else se2.insert(make_pair(ex/g,ey/g));
    }
    printf("%d\n",(int)se1.size() + (int)se2.size());
    return 0;
}

Codeforces Round #291 (Div. 2) B. Han Solo and Lazer Gun