数学水题

最新推荐文章于 2021-01-14 04:47:15 发布

原创最新推荐文章于 2021-01-14 04:47:15 发布 · 431 阅读

0 ·

CC 4.0 BY-SA版权

编程语言同时被 3 个专栏收录

307 篇文章

订阅专栏

==数学and思维==

131 篇文章

订阅专栏

codeforces

76 篇文章

订阅专栏

本文介绍了一个算法问题，即如何计算乘客能够乘坐到最近的 Beru 出租车所需的最短时间。该问题通过计算每辆出租车到达乘客位置所需的时间，并选取最小值来解决。

Beru-taxi

Time Limit:1000MS Memory Limit:262144KB 64bit IO Format:%I64d & %I64u

Submit Status Practice CodeForces 706A

Appoint description:

Description

Vasiliy lives at point (a, b) of the coordinate plane. He is hurrying up to work so he wants to get out of his house as soon as possible. New app suggested n available Beru-taxi nearby. The i-th taxi is located at point (x_i, y_i) and moves with a speed v_i.

Consider that each of n drivers will move directly to Vasiliy and with a maximum possible speed. Compute the minimum time when Vasiliy will get in any of Beru-taxi cars.

Input

The first line of the input contains two integers a and b ( - 100 ≤ a, b ≤ 100) — coordinates of Vasiliy's home.

The second line contains a single integer n (1 ≤ n ≤ 1000) — the number of available Beru-taxi cars nearby.

The i-th of the following n lines contains three integers x_i, y_i and v_i ( - 100 ≤ x_i, y_i ≤ 100, 1 ≤ v_i ≤ 100) — the coordinates of the i-th car and its speed.

It's allowed that several cars are located at the same point. Also, cars may be located at exactly the same point where Vasiliy lives.

Output

Print a single real value — the minimum time Vasiliy needs to get in any of the Beru-taxi cars. You answer will be considered correct if its absolute or relative error does not exceed 10^- 6.

Namely: let's assume that your answer is a, and the answer of the jury is b. The checker program will consider your answer correct, if $\text{[math]}$ .

Sample Input

Input

Output

1.00000000000000000000

Input

Output

0.50000000000000000000

#include<cstdio>
#include<cstring>
#include<algorithm>
#define M 0xffffff
#include<cmath>
using namespace std;
int x,y,v;
int main()
{
	int n,a,b;
	double s,t;
	while(~scanf("%d%d",&a,&b))
	{
		scanf("%d",&n);	
		t=M;
	    for(int i=1;i<=n;i++)
	{
	    scanf("%d%d%d",&x,&y,&v);
		s=sqrt((x-a)*(x-a)+(y-b)*(y-b))/v;
	     t=min(t,s);
	}
	printf("%.16lf\n",t);
	}
	return 0;
}