Codeforces Round #114 (Div. 1) A. Wizards and Trolleybuses 物理题
A. Wizards and Trolleybuses
Time Limit: 20 Sec
Memory Limit: 256 MB
题目连接
http://codeforces.com/contest/167/problem/A
Description
In some country live wizards. They love to ride trolleybuses.
A city in this country has a trolleybus depot with n trolleybuses. Every day the trolleybuses leave the depot, one by one and go to the final station. The final station is at a distance of d meters from the depot. We know for the i-th trolleybus that it leaves at the moment of time ti seconds, can go at a speed of no greater than vi meters per second, and accelerate with an acceleration no greater than ameters per second squared. A trolleybus can decelerate as quickly as you want (magic!). It can change its acceleration as fast as you want, as well. Note that the maximum acceleration is the same for all trolleys.
Despite the magic the trolleys are still powered by an electric circuit and cannot overtake each other (the wires are to blame, of course). If a trolleybus catches up with another one, they go together one right after the other until they arrive at the final station. Also, the drivers are driving so as to arrive at the final station as quickly as possible.
You, as head of the trolleybuses' fans' club, are to determine for each trolley the minimum time by which it can reach the final station. At the time of arrival at the destination station the trolleybus does not necessarily have zero speed. When a trolley is leaving the depot, its speed is considered equal to zero. From the point of view of physics, the trolleybuses can be considered as material points, and also we should ignore the impact on the speed of a trolley bus by everything, except for the acceleration and deceleration provided by the engine.
Input
The first input line contains three space-separated integers n, a, d (1 ≤ n ≤ 105, 1 ≤ a, d ≤ 106) — the number of trolleybuses, their maximum acceleration and the distance from the depot to the final station, correspondingly.
Next n lines contain pairs of integers ti vi (0 ≤ t1 < t2... < tn - 1 < tn ≤ 106, 1 ≤ vi ≤ 106) — the time when the i-th trolleybus leaves the depot and its maximum speed, correspondingly. The numbers in the lines are separated by spaces.
Output
For each trolleybus print a single line the time it arrives to the final station. Print the times for the trolleybuses in the order in which the trolleybuses are given in the input. The answer will be accepted if the absolute or relative error doesn't exceed 10 - 4.
Sample Input
- 3 10 10000
0 10
5 11
1000 1
Sample Output
- 1000.5000000000
1000.5000000000
11000.0500000000
HINT
题意
有n辆车,车的加速度为a,他们都要从起点开到终点,起点终点距离为d
第i辆车的发车时间是t[i],最大速度是v[i]
保证,不能超车
问你每辆车到达终点的时间是多少
题解:
高中物理题,稍微推推公式O1回答就好了
代码:
- #include<iostream>
- #include<stdio.h>
- #include<math.h>
- using namespace std;
- int n;
- double a,d;
- double t[],v[];
- double ans[];
- int main()
- {
- scanf("%d%lf%lf",&n,&a,&d);
- for(int i=;i<n;i++)
- {
- scanf("%lf%lf",&t[i],&v[i]);
- if(v[i]*v[i]/(*a)>=d)
- ans[i]=sqrt(.*d/a)+t[i];
- else
- ans[i]=v[i]/a + (d-v[i]*v[i]/(2.0*a))/v[i]+t[i];
- }
- for(int i=;i<n;i++)
- {
- if(ans[i]<ans[i-])
- ans[i]=ans[i-];
- }
- for(int i=;i<n;i++)
- printf("%.10f\n",ans[i]);
- }
Codeforces Round #114 (Div. 1) A. Wizards and Trolleybuses 物理题的更多相关文章
- Codeforces Round #114 (Div. 1) B. Wizards and Huge Prize 概率dp
B. Wizards and Huge Prize Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest ...
- Codeforces Round #114 (Div. 1) D. Wizards and Roads 笛卡尔树+树贪心+阅读题
D. Wizards and Roads 题目连接: http://www.codeforces.com/contest/167/problem/D Description In some count ...
- Codeforces Round #114 (Div. 1) E. Wizards and Bets 高斯消元
E. Wizards and Bets 题目连接: http://www.codeforces.com/contest/167/problem/E Description In some countr ...
- Codeforces Round #114 (Div. 1) C. Wizards and Numbers 博弈论
C. Wizards and Numbers 题目连接: http://codeforces.com/problemset/problem/167/C Description In some coun ...
- Codeforces Round #114 (Div. 2)
Codeforces Round #114 (Div. 2) 代码 Codeforces Round #114 (Div. 2) C. Wizards and Trolleybuses 思路 每条车的 ...
- Codeforces Round #297 (Div. 2)A. Vitaliy and Pie 水题
Codeforces Round #297 (Div. 2)A. Vitaliy and Pie Time Limit: 2 Sec Memory Limit: 256 MBSubmit: xxx ...
- Codeforces Round #298 (Div. 2) A、B、C题
题目链接:Codeforces Round #298 (Div. 2) A. Exam An exam for n students will take place in a long and nar ...
- Codeforces Round #327 (Div. 2) A. Wizards' Duel 水题
A. Wizards' Duel Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/591/prob ...
- Codeforces Round #384 (Div. 2) C. Vladik and fractions 构造题
C. Vladik and fractions 题目链接 http://codeforces.com/contest/743/problem/C 题面 Vladik and Chloe decided ...
随机推荐
- Spring学习之AOP
Spring-AOP(Aspect-orented programming) 在业务流程中插入与业务无关的逻辑,这样的逻辑称为Cross-cutting concerns,将Crossing-cutt ...
- Android 如何使用juv-rtmp-client.jar向Red5服务器发布实时视频数据
使用juv-client-client.jar主要是尽快地完成毕业设计里面手机端向网页端发送实时视频的功能,由于实习和做毕业设计的时间冲突,因此完成毕业设计只花了1个多月时间. (万恶的形式主义,论文 ...
- 通过userAgent判断手机浏览器类型
我们可以通过userAgent来判断,比如检测某些关键字,例如:AppleWebKit*****Mobile或AppleWebKit,需要注意的是有些浏览器的userAgent中并不包含AppleWe ...
- ANDROID开发之SQLite详解
本文转自:http://www.cnblogs.com/Excellent/archive/2011/11/19/2254888.html
- [Papers]NSE, $\pi$, Lorentz space [Suzuki, JMFM, 2012]
$$\bex \sen{\pi}_{L^{s,\infty}(0,T;L^{q,\infty}(\bbR^3))} \leq \ve_*, \eex$$ with $$\bex \frac{2}{s} ...
- 认识Java虚拟机的内部体系结构、gc示例
认识Java虚拟机的内部体系结构 Java虚拟机的内部体系结构也许很少有人去关心,因为对于Java程序员来说,一般只需要跟API打交道就可以了.这些体系结构只是Java虚拟机内部的结构而已.但是如果理 ...
- 转载:fstream和ifstream详细用法
文件 I/O 在C++中比烤蛋糕简单多了.在这篇文章里,我会详细解释ASCII和二进制文件的输入输出的每个细节,值得注意的是,所有这些都是用C++完成的. 一.ASCII 输出 为了使用下面的方法, ...
- bug报告-常用词汇中英对照表
- PHP 实现下载文件的方法
方法一: header('Content-Description: File Transfer'); header('Content-Type: application/octet-stream'); ...
- DateTime日期计算
//今天 DateTime.Now.Date.ToShortDateString(); //昨天,就是今天的日期减一 DateTime.Now.AddDays(-1).ToShortDateStrin ...