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
- <五>面向对象分析之UML核心元素之边界
一:基本概念
- 【转】visual studio 2012进行C语言开发[图文]
原文网址:http://blog.csdn.net/chengyafei0104/article/details/9826025 现在大家计算机大概都脱离XP了,so,之前蛮多可以用的编译器,可能放在 ...
- Web开发人员必备工具-Emmet (Zen Coding)
如果你从事前端开发或者web开发的话,一定听说过Zen coding - 一种快速编写HTML/CSS代码的方法.它使用仿CSS选择器的语法来快速开发HTML和CSS - 由Sergey Chikuy ...
- MyBatis学习 之 二、SQL语句映射文件(1)resultMap
目录(?)[-] 二SQL语句映射文件1resultMap resultMap idresult constructor association联合 使用select实现联合 使用resultMap实 ...
- history命令
在 Linux 下面可以使用 history 命令查看用户的所有历史操作,同时 shell 命令操作记录默认保存在用户目录的 .bash_history 文件中.通过这个文件可以查询 shell 命令 ...
- XTUOJ 1248 TC or CF 搜索
这个题一眼看上去不会 然后有人说是网络流 然后我就想怎么建图啊,然后不会(是本蒟蒻太垃圾了),肯定有网络流解法 然后去群里问了gdut的巨巨,他说他队友爆搜+剪枝过了(我也是非常的叹服) 然后我也写了 ...
- List转换成Json、对象集合转换Json等
#region List转换成Json /// <summary> /// List转换成Json /// </summary> public static string Li ...
- 重新执笔,已是大三!Jekyll自定义主题开发
前言 “一转眼忘了时间 丢了感觉 黑了世界 再逞强 再疯狂 也会伤 不知 不觉 后知 后觉 然后 发现 失去 知觉 ”——<一吻不天荒> 感言 时间是把双刃剑,什么解决不了,忧烦的,慢慢变 ...
- 简单的flash策略文件服务器!
最近在做一个flash小游戏,众所周知,flash连接服务器socket的时候,需要向服务器请求策略文件.以下是一个简单的策略文件服务器的代码c++: #include <Winsock2.h& ...
- [转]Erlang不能错过的盛宴
Erlang不能错过的盛宴 (快步进入Erlang的世界) 作者:成立涛 (litaocheng@gmail.com) 作为程序员,我们曾经闻听很多“业界动态”,“技术革新”,曾经接触很多“高手箴言” ...