B. Han Solo and Lazer Gun

最新推荐文章于 2019-08-13 08:57:03 发布

原创最新推荐文章于 2019-08-13 08:57:03 发布 · 489 阅读

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CC 4.0 BY-SA版权

本文介绍了一种算法，用于计算在给定武器位置和敌人坐标的情况下，消灭所有敌人的最少射击方向数。通过确定不同斜率的数量来解决此问题。

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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

Output

Input

2 1 2
1 1
1 0

Output

Note

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

$\text{[math]}$

题意：给出武器位置和敌人位置，要求计算武器小灭所有敌人所需的最少直线方向，其实就是求坐标系中所有敌人组成的不同斜率有多少；

#include<cstring>
#include<cstdio>
#include<iostream>
#include<set>

using namespace std;

int main()
{
    int n;
    double xx,yy;
    while(cin>>n>>xx>>yy)
    {
        int p=0;
        set<double>s;
        for(int i=0;i<n;i++)
        {
            double a,b;
            cin>>a>>b;
            if(fabs(a-xx)<=1e-10) p=1;
            else s.insert((b-yy)/(a-xx));
        }
        cout<<s.size()+p<<endl;
    }
}