Osu!

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)

Total Submission(s): 58    Accepted Submission(s): 41

Special Judge

Problem Description
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 ti at place (xi, yi), and you should click it exactly at ti at (xi, yi). 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 ti and ti+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.
 
Input
The first line contains an integer T (T ≤ 10), denoting the number of the test cases.



For each test case, the first line contains an integer N (2 ≤ N ≤ 1000) denoting the number of the points in the game.  Then N lines follow, the i-th line consisting of 3 space-separated integers, ti(0 ≤ ti < ti+1 ≤ 106),
xi, and yi (0 ≤ xi, yi ≤ 106) as mentioned above.
 
Output
For each test case, output the answer in one line.



Your answer will be considered correct if and only if its absolute or relative error is less than 1e-9.
 
Sample Input
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
 
Sample Output
9.2195444573
54.5893762558
鞍山现场赛第签到题,超级水!!
求最大难度,难度为相邻两点的距离除以时间差。
#include <stdio.h>
#include <string.h>
#include <iostream>
#include <math.h>
#include <algorithm>
#include <vector>
#include <map>
#include <queue>
#include <stack>
#define PI acos(-1.0)
#define eps 1e-8
#define LL long long
#define moo 1000000007
#define INF -999999999
using namespace std;
long long a[1005],b[1005],c[1005];
double d[1005];
int main()
{
int t;
int n;
scanf("%d",&t);
while(t--)
{ scanf("%d",&n);
double ans=0;
for(int i=0;i<n;i++)
{
scanf("%I64d%I64d%I64d",&a[i],&b[i],&c[i]);
}
for(int i=1;i<n;i++)
{
d[i]=sqrt((b[i]-b[i-1])*(b[i]-b[i-1])+(c[i]-c[i-1])*(c[i]-c[i-1]))/(a[i]-a[i-1]);
ans=max(d[i],ans);
}
printf("%.10f\n",ans);
}
}

hdu 5078的更多相关文章

  1. 2014 Asia AnShan Regional Contest --- HDU 5078 Osu!

    Osu! Problem's Link:   http://acm.hdu.edu.cn/showproblem.php?pid=5078 Mean: 略. analyse: 签到题,直接扫一遍就得答 ...

  2. hdu 5078 Osu! (2014 acm 亚洲区域赛鞍山 I)

    题目链接:http://acm.hdu.edu.cn/showproblem.php? pid=5078 Osu! Time Limit: 2000/1000 MS (Java/Others)     ...

  3. hdu 5078 2014鞍山现场赛 水题

    http://acm.hdu.edu.cn/showproblem.php?pid=5078 现场最水的一道题 连排序都不用,由于说了ti<ti+1 //#pragma comment(link ...

  4. HDU - 5078 水题

    题意:最大困难=距离 / 相邻时间 #include<cstring> #include<cstdio> #include<cmath> #define ll do ...

  5. hdu 5078(2014鞍山现场赛 I题)

    数据 表示每次到达某个位置的坐标和时间 计算出每对相邻点之间转移的速度(两点间距离距离/相隔时间) 输出最大值 Sample Input252 1 9//t x y3 7 25 9 06 6 37 6 ...

  6. [ACM] HDU 5078 Osu!

    Osu! Problem Description Osu! is a very popular music game. Basically, it is a game about clicking. ...

  7. HDU 5078 Revenge of LIS II(dp LIS)

    Problem Description In computer science, the longest increasing subsequence problem is to find a sub ...

  8. hdu 5078 Osu!(鞍山现场赛)

    Osu! Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others) Total Sub ...

  9. 2014 ACM-ICPC Asia Anshan Regional Contest(Online Version)

    题目I - Osu! - HDU 5078 题目分析:最水的一道题吧,求两点间的距离和时间差值的最大比值 #include<stdio.h> #include<math.h> ...

随机推荐

  1. phpMailer中文说明[转]

    A开头: $AltBody--属性出自:PHPMailer::$AltBody文件:class.phpmailer.php说明:该属性的设置是在邮件正文不支持HTML的备用显示 AddAddress- ...

  2. Python的多进程编程

    Python在2.6引入了多进程的机制,并提供了丰富的组件及api以方便编写并发应用.multiprocessing包的组件Process, Queue, Pipe, Lock等组件提供了与多线程类似 ...

  3. 怎样推断多个字段组成的keyword在另外一张表中是否存在

    怎样推断多个字段组成的keyword在另外一张表中是否存在 老帅(20141107) 1.首先推断一个keyword在另外一张表中是否存在非常easy! SELECT * FROM a WHERE a ...

  4. 强大的vim配置文件,让编程更随意(转)

    欢迎来到小码哥的博客 博客搬家啦 blog.ma6174.com 强大的vim配置文件,让编程更随意 花了很长时间整理的,感觉用起来很方便,共享一下. 我的vim配置主要有以下优点: 1.按F5可以直 ...

  5. 安卓Camera APP

    一.Camera package android.hardware            该类用于设定图像捕获设置,开启/关闭预览,抓拍图片以及获取帧用于编码视频.这个类是Camera服务的客户端,用 ...

  6. CAS 与 无锁队列

    http://coolshell.cn/articles/8239.html http://www.tuicool.com/articles/VZ3IBv http://blog.csdn.net/r ...

  7. 【Windows】免费图片提取文字的方法

    今天意外的看到一个可以提取图片中文字的网站,自己试了下,提取效果还不错 网址为: https://zhcn.109876543210.com/ 现在有图片如下 我想从中提取的文字 1.打开网址,上传图 ...

  8. 启动loadrunner 11的controller提示试图执行系统不支持的操作(已解决)

    启动loadrunner 11的controller提示试图执行系统不支持的操作 分类: loadrunner 2014-05-12 17:33 532人阅读 评论(0) 收藏 举报 win7旗舰版3 ...

  9. Linux-Nginx-关闭进程

    当然就仅仅是介绍一条命令了,就这么简单. nginx默认创建一个工作进程 root 2713 1 0 07:56 ? 00:00:00 nginx: master process ../sbin/ng ...

  10. 【Shiro】Apache Shiro架构之权限认证(Authorization)

    Shiro系列文章: [Shiro]Apache Shiro架构之身份认证(Authentication) [Shiro]Apache Shiro架构之集成web [Shiro]Apache Shir ...