题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=5078

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.
 

鞍山的签到题，记得把类型定义成long long 防止溢出就行了。

#include<iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<map>
#include<set>
#include<vector>
#include<cstdlib>
using namespace std;
#define CLR(A) memset(A,0,sizeof(A))
typedef long long ll;
const int MAXN=1010;
ll x[MAXN],y[MAXN],t[MAXN];
int main(){
    int T;
    cin>>T;
    while(T--){
        int n;
        cin>>n;
        for(int i=0;i<n;i++) cin>>t[i]>>x[i]>>y[i];
        double ans=0;
        for(int i=1;i<n;i++){
            double tmp=sqrt((x[i]-x[i-1])*(x[i]-x[i-1])+(y[i]-y[i-1])*(y[i]-y[i-1]))/(double)(t[i]-t[i-1]);
            ans=max(ans,tmp);
        }
        printf("%.10lf\n",ans);
    }
    return 0;
}

hdu 5078 Osu! （2014 acm 亚洲区域赛鞍山 I）

原文：http://blog.csdn.net/asdfghjkl1993/article/details/40398675