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hdu 5078 Osu! (2014 acm 亚洲区域赛鞍山 I)

时间:2017-06-05 21:05:18      阅读:261      评论:0      收藏:0      [点我收藏+]

标签:space   pid   eof   set   discus   note   mod   hint   man   

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.

技术分享

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
 

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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\n",ans);
    }
    return 0;
}




hdu 5078 Osu! (2014 acm 亚洲区域赛鞍山 I)

标签:space   pid   eof   set   discus   note   mod   hint   man   

原文地址:http://www.cnblogs.com/zsychanpin/p/6946950.html

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