zoukankan      html  css  js  c++  java
  • 2014ACM/ICPC亚洲区鞍山赛区现场赛1009Osu!

    鞍山的签到题,求两点之间的距离除以时间的最大值。直接暴力过的。

    A - Osu!
    Time Limit:1000MS     Memory Limit:262144KB     64bit IO Format:%I64d & %I64u
    Appoint description: 

    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 t i at place (x i, y i), and you should click it exactly at t iat (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.
     

    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, t i(0 ≤ t i < t i+1 ≤ 10 6), x i, and y i (0 ≤ x i, y i ≤ 10 6) 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

    Hint

    In memory of the best osu! player ever Cookiezi.
     1 #include<cstdio>
     2 #include<algorithm>
     3 #include<cmath>
     4 #include<stdlib.h>
     5 using namespace std;
     6 const int maxn=1000+10;
     7 struct
     8 {
     9     double t;
    10     double x;
    11     double y
    12     ;
    13 }Node[maxn];
    14 int main()
    15 {
    16     int T;
    17     scanf("%d",&T);
    18     while(T--)
    19     {
    20         int N;
    21         double maxd=-1;
    22         scanf("%d",&N);
    23         for(int i=0;i<N;i++)
    24             scanf("%lf%lf%lf",&Node[i].t,&Node[i].x,&Node[i].y);
    25         for(int i=0;i<N;i++)
    26         {
    27             for(int j=i+1;j<N;j++)
    28             {
    29                 double time=sqrt((Node[i].x-Node[j].x)*(Node[i].x-Node[j].x)+(Node[i].y-Node[j].y)*(Node[i].y-Node[j].y))/fabs(Node[i].t-Node[j].t);
    30                 maxd=max(maxd,time);
    31             }
    32         }
    33         printf("%.10lf
    ",maxd);
    34     }
    35     return 0;
    36 }
  • 相关阅读:
    生物神经网络和人工神经网络浅谈
    卷积神经网络
    DOM进阶之HTMl属性操作(对象属性)
    01 selenium基本使用补充
    01 selenium基本使用
    day4笔记
    03 获取豆瓣电影top250
    02 爬取视频
    day3笔记
    01 requests基本使用
  • 原文地址:https://www.cnblogs.com/wpnan/p/4043880.html
Copyright © 2011-2022 走看看