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 nad (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

  1. 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 ...

  2. 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 ...

  3. 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 ...

  4. 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 ...

  5. 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 ...

  6. 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  ...

  7. 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 ...

  8. 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 ...

  9. 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

  1. ecshop 二次开发及模板标签

    ecs_account_log // 用户账目日志表   ecs_activity // 活动表(代码,名称,开始,结束,描述)   ecs_ad // 广告表(位置,类型,名称,链接,图片,开始,结 ...

  2. C 的 coroutine 库 via 云风的 BLOG

    今天实现了一个 C 用的 coroutine 库. 我相信这个东西已经被无数 C 程序员实现过了, 但是通过 google 找了许多, 或是接口不让我满意, 或是过于重量. 在 Windows 下, ...

  3. [Papers]NSE, $u_3$, Lebesgue space [Zhou-Pokorny, Nonlinearity, 2009]

    $$\bex u_3\in L^p(0,T;L^q(\bbR^3)),\quad \frac{2}{p}+\frac{3}{q}=\frac{3}{4}+\frac{1}{2q},\quad \fra ...

  4. 数组乘积--满足result[i] = input数组中除了input[i]之外所有数的乘积(假设不会溢出

    数组乘积(15分) 输入:一个长度为n的整数数组input 输出:一个长度为n的整数数组result,满足result[i] = input数组中除了input[i]之外所有数的乘积(假设不会溢出). ...

  5. Oracle 客户端安装 + pl/sql工具安装配置

    Oracle 客户端安装 +  pl/sql工具安装配置 下载oracle客户端,并在本地安装. 11g下载地址为: http://www.oracle.com/technetwork/databas ...

  6. bjfu1100 圆环

    这题也是2011百度之星的一道题.知道做法后代码极简单. 不过我做完后随便上网搜了一下,发现竟然还有很多不同的做法.别的做法我就不管了,我只把我的做法的原理说清楚.我做题时是按如下顺序逐步找到规律的: ...

  7. winscp配置

    WinSCP Install and run WinSCP Go to Preferences (Ctrl+Alt+P) and click on Transfer, then on Add. Nam ...

  8. 内核源码分析之进程地址空间(基于3.16-rc4)

    所谓进程的地址空间,指的就是进程的虚拟地址空间.当创建一个进程时,内核会为该进程分配一个线性的地址空间(虚拟地址空间),有了虚拟地址空间后,内核就可以通过页表将进程的物理地址地址空间映射到其虚拟地址空 ...

  9. iOS开发中使用Bmob RESTful API

    简介 尽管Bmob已经提供了一套SDK供开发者使用,但有时候开发者可能希望能直接与Bmob后台进行直接交互,以达到某些特别的需求(直接操作_User表.同步网络请求等).而RESTful API可以使 ...

  10. elasticsearch配置文件解析

    配置es的集群名称 : cluster.name:  fcz_es