CF291-B

B. Han Solo and Lazer Gun

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

There are n Imperial stormtroopers on the field. The battle field is a plane with Cartesian coordinate system. Each stormtrooper is associated with his coordinates (x, y) on this plane.

Han Solo has the newest duplex lazer gun to fight these stormtroopers. It is situated at the point (x₀, y₀). In one shot it can can destroy all the stormtroopers, situated on some line that crosses point (x₀, y₀).

Your task is to determine what minimum number of shots Han Solo needs to defeat all the stormtroopers.

The gun is the newest invention, it shoots very quickly and even after a very large number of shots the stormtroopers don't have enough time to realize what's happening and change their location.

Input

The first line contains three integers n, x₀ и y₀ (1 ≤ n ≤ 1000,  - 10⁴ ≤ x₀, y₀ ≤ 10⁴) — the number of stormtroopers on the battle field and the coordinates of your gun.

Next n lines contain two integers each x_i, y_i ( - 10⁴ ≤ x_i, y_i ≤ 10⁴) — the coordinates of the stormtroopers on the battlefield. It is guaranteed that no stormtrooper stands at the same point with the gun. Multiple stormtroopers can stand at the same point.

Output

Print a single integer — the minimum number of shots Han Solo needs to destroy all the stormtroopers.

Sample test(s)

input

4 0 0
1 1
2 2
2 0
-1 -1

output

2

input

2 1 2
1 1
1 0

output

1

Note

Explanation to the first and second samples from the statement, respectively:

 给定n个敌人的位置和枪的起始位置,一枪可以射杀经过枪位置直线上的所有敌人

问至少需要多少枪可以杀光所有敌人

数学题,用斜率或者叉积都可以求解

我用的是斜率,将枪的位置看成原点,求所有敌人相对枪的坐标,则敌人i的斜率为yi/xi

枚举每个活着敌人,将和该敌人相同斜率的人都杀光,子弹数+1

遍历完后子弹数即为答案

#include <iostream>

using namespace std;

int main()

{

    int n,x,y;

    int use[1010];

    double px[1010];

    double py[1010];

    cin>>n>>x>>y;

    for(int i=1;i<=n;i++)

    {

        cin>>px[i]>>py[i];

        px[i]-=x;

        py[i]-=y;

    }

    int res=0;

    bool flag=false;

    for(int i=1;i<=n;i++)

    {

        if(px[i]==0)

        {

            use[i]=1;

            if(!flag)

            {

                flag=true;

                res++;

            }

            else

                continue;

        }

        if(!use[i])

        {

            res++;

            use[i]=1;

            double k=py[i]/px[i];

            for(int j=1;j<=n;j++)

            {

                if(py[j]/px[j]==k&&!use[j])

                    use[j]=1;

            }

        }

    }

    cout<<res<<endl;

    return 0;

}