hdu 5112 A Curious Matt
题目连接
http://acm.hdu.edu.cn/showproblem.php?pid=5112
A Curious Matt
Description
There is a curious man called Matt.
One day, Matt's best friend Ted is wandering on the non-negative half of the number line. Matt finds it interesting to know the maximal speed Ted may reach. In order to do so, Matt takes records of Ted’s position. Now Matt has a great deal of records. Please help him to find out the maximal speed Ted may reach, assuming Ted moves with a constant speed between two consecutive records.
Input
The first line contains only one integer $T$, which indicates the number of test cases.
For each test case, the first line contains an integer $N\ (2 \leq N \leq 10000)$,indicating the number of records.
Each of the following $N$ lines contains two integers $t_i$ and $x_i$ $(0 \leq t_i, x_i \leq 10^6)$, indicating the time when this record is taken and Ted’s corresponding position. Note that records may be unsorted by time. It’s guaranteed that all ti would be distinct.
Output
For each test case, output a single line “Case #x: y”, where x is the case number (starting from 1), and y is the maximal speed Ted may reach. The result should be rounded to two decimal places.
Sample Input
2
3
2 2
1 1
3 4
3
0 3
1 5
2 0
Sample Output
Case #1: 2.00
Case #2: 5.00
简单题,权当练习一下英文。。
#include<algorithm>
#include<iostream>
#include<cstdlib>
#include<cstring>
#include<cstdio>
#include<vector>
#include<map>
using std::max;
using std::cin;
using std::cout;
using std::endl;
using std::fabs;
using std::sort;
using std::pair;
using std::vector;
#define pb(e) push_back(e)
#define sz(c) (int)(c).size()
#define mp(a, b) make_pair(a, b)
#define all(c) (c).begin(), (c).end()
#define iter(c) decltype((c).begin())
#define cls(arr,val) memset(arr,val,sizeof(arr))
#define cpresent(c, e) (find(all(c), (e)) != (c).end())
#define rep(i, n) for (int i = 0; i < (int)(n); i++)
#define tr(c, i) for (iter(c) i = (c).begin(); i != (c).end(); ++i)
const int N = ;
typedef unsigned long long ull;
struct Node {
int t, x;
bool operator<(const Node &b) const {
return t < b.t;
}
}rec[N];
int main() {
#ifdef LOCAL
freopen("in.txt", "r", stdin);
freopen("out.txt", "w+", stdout);
#endif
int t, n, k = ;
scanf("%d", &t);
while (t--) {
double ans = 0.0;
scanf("%d", &n);
rep(i, n) scanf("%d %d", &rec[i].t, &rec[i].x);
sort(rec, rec + n);
for (int i = ; i < n; i++) {
ans = max(ans, fabs((double)rec[i].x - rec[i - ].x) / (rec[i].t - rec[i - ].t));
}
printf("Case #%d: %.2lf\n", k++, ans);
}
return ;
}
hdu 5112 A Curious Matt的更多相关文章
- HDU 5112 A Curious Matt 水题
A Curious Matt Time Limit: 1 Sec Memory Limit: 256 MB 题目连接 http://acm.hdu.edu.cn/showproblem.php?pid ...
- HDU 5112 A Curious Matt (2014ACM/ICPC亚洲区北京站-重现赛)
A Curious Matt Time Limit: 2000/2000 MS (Java/Others) Memory Limit: 512000/512000 K (Java/Others) ...
- 水题:HDU 5112 A Curious Matt
Description There is a curious man called Matt. One day, Matt's best friend Ted is wandering on the ...
- hdoj 5112 A Curious Matt
A Curious Matt Time Limit: 2000/2000 MS (Java/Others) Memory Limit: 512000/512000 K (Java/Others) ...
- HDU 5120 A Curious Matt(2014北京赛区现场赛A题 简单模拟)
题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=5112 解题报告:扫一遍 #include<cstdio> #include<cstr ...
- hdu 5112 (2014北京 水)
题意:有个人在跑步,一直每个时间他所在的位置,求最大速度 #include <iostream> #include <cstring> #include <cstdio& ...
- hdu 5112 (2014北京现场赛 A题)
给出某个时刻对应的速度 求出相邻时刻的平均速度 输出最大值 Sample Input23 // n2 2 //t v1 13 430 31 52 0 Sample OutputCase #1: 2.0 ...
- HDU 5119 Happy Matt Friends (背包DP + 滚动数组)
题目链接:HDU 5119 Problem Description Matt has N friends. They are playing a game together. Each of Matt ...
- 2014年亚洲区域赛北京赛区现场赛A,D,H,I,K题解(hdu5112,5115,5119,5220,5122)
转载请注明出处: http://www.cnblogs.com/fraud/ ——by fraud 下午在HDU上打了一下今年北京区域赛的重现,过了5题,看来单挑只能拿拿铜牌,呜呜. ...
随机推荐
- Eclipse SVN冲突解决
基本原则是:每次提交前需要先和线上的对比,先把冲突解决掉,然后把线上的更新到本地,最后把本地的提交上去. 右键项目 -> Team -> 与资源库同步 在同步视图中选择Conflicts ...
- 学习记录 Eclipse常用快捷键及其演练
Eclipse中10个最有用的快捷键组合 1. ctrl+shift+r:打开资源 这可能是所有快捷键组合中最省时间的了.这组快捷键可以让你打开你的工作区中任何一个文件,而你只需要按下文件名或mask ...
- php文件大小单位转换GB MB KB
private function formatBytes($size){ $units = array('字节','K','M','G','T'); $i = 0; for( ; $size>= ...
- php json中文处理方法,请json更懂中文
1.php5.3版本及以下.的处理方式 /** *php5.3版本以前,json中文问题的解决解决方案 */ function encode_json($str) { return urldecode ...
- leetcode 112
112. Path Sum Given a binary tree and a sum, determine if the tree has a root-to-leaf path such that ...
- 必须会的SQL语句(一) 创建数据库与删除数据库
1.创建数据库 Create database 名称 on primary { name ='名称', filename ='c:\xx\名称.mdf', size = 10mb ...
- 介绍一点.NET反编译的知识
反编译是我们理解.NET内部实现的一种良好的手段. 程序编译时 Test.exe是IL代码.我们可以通过一些工具,来查看这些IL代码. 一模一样? 理论上来说,一模一样的反编译是不存在的.原因有以下3 ...
- 【风马一族_xml】Schema--- xml的约束技术
Schema 是xml的约束技术,出现的目的是为了替代dtd 本身也是一个xml,非常方便使用xml的解析引擎进行解析 对名称空间有非常好的支持 支持更多的数据类型,并且支持用户自定义数据类型 可以进 ...
- 开篇 hello 内Cool超人
经过一年时间看到asp.net mvc一直被受微软开发团队的注重.与之相比的silverlight我感觉到有点力不从心.除去silverlight第一次运行要安装Runtime不说,产品不可能只运行在 ...
- js 实现栈
function Stack() { this.dataStore = []; this.top = 0; this.push=push; this.pop=pop; this.peek=peek; ...