zoukankan      html  css  js  c++  java
  • 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)    Memory Limit: 262144/262144 K (Java/Others)
    Total Submission(s): 180    Accepted Submission(s): 114
    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
    Hint
    In memory of the best osu! player ever Cookiezi.
     

    Source
     

    Recommend
    liuyiding   |   We have carefully selected several similar problems for you:  5081 5080 5079 5077 5076 
     

    Statistic | Submit | Discuss | Note


    鞍山的签到题,记得把类型定义成long long 防止溢出即可了。


    #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
    ",ans);
        }
        return 0;
    }
    




  • 相关阅读:
    LeetCode OJ Remove Duplicates from Sorted Array II
    LeetCode OJ 75. Sort Colors
    LeetCode OJ 74. Search a 2D Matrix
    LeetCode OJ 73. Set Matrix Zeroes
    Taglist
    NERDTree
    Beyond Compare 4
    Beyond compare vs kdiff3
    切換 java compiler 版本
    PE+ 1.0 ( Pump Express Plus 1.0 )
  • 原文地址:https://www.cnblogs.com/zsychanpin/p/6946950.html
Copyright © 2011-2022 走看看