Osu!

Osu! is a very popular music game. Basically, it is a game about clicking. Some points will appear on the screen at some time, and you have to click them at a correct time.






Now, you want to write an algorithm to estimate how diffecult a game is.



To simplify the things, in a game consisting of N points, point i will occur at time t_i at place (x_i, y_i), and you should click it exactly at t_i at (x_i, y_i). That means you should move your cursor
from point i to point i+1. This movement is called a jump, and the difficulty of a jump is just the distance between point i and point i+1 divided by the time between t_i and t_i+1. And the difficulty of a game is simply the difficulty
of the most difficult jump in the game.



Now, given a description of a game, please calculate its difficulty.

2

5

2 1 9

3 7 2

5 9 0

6 6 3

7 6 0

10

11 35 67

23 2 29

29 58 22

30 67 69

36 56 93

62 42 11

67 73 29

68 19 21

72 37 84

82 24 98

9.2195444573

54.5893762558

Hint

In memory of the best osu! player ever Cookiezi.

解题思路：

水题，看懂题意，写代码就没问题。

代码：

#include <iostream>

#include <stdio.h>

#include <algorithm>

#include <string.h>

#include <cmath>

#include <iomanip>

#include <vector>

#include <map>

#include <stack>

#include <queue>

using namespace std;

int n;

struct Point

{

    int x,y,t;

}point[1002];

double dis(Point a,Point b)

{

    return sqrt((double)(a.x-b.x)*(a.x-b.x)+(double)(a.y-b.y)*(a.y-b.y));

}

int main()

{

    int t;cin>>t;

    while(t--)

    {

        cin>>n;

        cin>>point[1].t>>point[1].x>>point[1].y;

        double ans=-1;

        for(int i=2;i<=n;i++)

        {

            cin>>point[i].t>>point[i].x>>point[i].y;

            double temp=dis(point[i],point[i-1])/(point[i].t-point[i-1].t);

            if(ans<temp)

                ans=temp;

        }

        cout<<setiosflags(ios::fixed)<<setprecision(9)<<ans<<endl;

    }

    return 0;

}