Codeforces 514 B . Han Solo and Lazer Gun 精度除0

最新推荐文章于 2020-06-05 22:57:20 发布

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最新推荐文章于 2020-06-05 22:57:20 发布

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文章标签： Codeforces

本文链接：https://blog.youkuaiyun.com/haoliang94/article/details/43890147

本文解析了CodeForces竞赛中B题“HanSoloandLazerGun”的解决方案，该题涉及确定使用激光枪消灭所有战场上的敌人的最少射击次数。通过坐标计算与排序算法，文章详细介绍了如何高效地解决此问题。

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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 xi, yi ( - 104 ≤ xi, yi ≤ 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

$\text{[math]}$

题目分析。。。eps别忘记了。。还有除0的特殊判断。。脑残了半天，除0没有判断出来

代码：

#include<iostream>
#include<string>
#include<algorithm>
#include<cstdlib>
#include<cstdio>
#include<set>
#include<map>
#include<vector>
#include<cstring>
#include<stack>
#include<cmath>
#include<queue>
#define INF 0x0f0f0f0f
#define eps 1e-10
using namespace std;
struct Point
{
	double x;
	double y;
};

int main()
{
	int i,j,k,l,m,n,sum=1,flag=0;
	double x,y;
	double tana[1005];
	Point point[1005];
	scanf("%d%lf%lf",&n,&x,&y);
	for(i=0;i<n;i++)
	{
		scanf("%lf%lf",&point[i].x,&point[i].y);
	}
	for(i=0;i<n;i++)
	{
		if(fabs(point[i].x-x)>eps)
		tana[i]=(point[i].y-y)*1.0/(point[i].x-x);
		else
		{
			tana[i]=1000000000000;
		}
	}
	sort(tana,tana+n);
	for(i=1;i<n;i++)
	{
		//printf("%.2lf ",tana[i]);
		if(tana[i]-tana[i-1]>eps)	sum++;
	}
	printf("%d\n",sum);
	
	

}