zoukankan      html  css  js  c++  java
  • Codeforces Beta Round #5 D. Follow Traffic Rules 物理

    D. Follow Traffic Rules

    题目连接:

    http://www.codeforces.com/contest/5/problem/D

    Description

    Everybody knows that the capital of Berland is connected to Bercouver (the Olympic capital) by a direct road. To improve the road's traffic capacity, there was placed just one traffic sign, limiting the maximum speed. Traffic signs in Berland are a bit peculiar, because they limit the speed only at that point on the road where they are placed. Right after passing the sign it is allowed to drive at any speed.

    It is known that the car of an average Berland citizen has the acceleration (deceleration) speed of a km/h2, and has maximum speed of v km/h. The road has the length of l km, and the speed sign, limiting the speed to w km/h, is placed d km (1 ≤ d < l) away from the capital of Berland. The car has a zero speed at the beginning of the journey. Find the minimum time that an average Berland citizen will need to get from the capital to Bercouver, if he drives at the optimal speed.

    The car can enter Bercouver at any speed.

    Input

    The first line of the input file contains two integer numbers a and v (1 ≤ a, v ≤ 10000). The second line contains three integer numbers l, d and w (2 ≤ l ≤ 10000; 1 ≤ d < l; 1 ≤ w ≤ 10000).

    Output

    Print the answer with at least five digits after the decimal point.

    Sample Input

    1 1
    2 1 3

    Sample Output

    2.500000000000

    Hint

    题意

    有一个长度为l的道路,你的加速是a

    从[0,d]的限速是w,[d,l]的限速是v,问你最少花费多少时间从起点到终点。

    题解:

    高中物理题,四种情况,都讨论一下,然后瞎搞搞

    代码

    #include<bits/stdc++.h>
    using namespace std;
    
    double a,v,l,d,w;
    int main()
    {
        cin>>a>>v>>l>>d>>w;
        double t = w/a;
        w = min(min(w, v), sqrt(2 * a * d));
        if (2*a*d+w*w>2*v*v)
            t = (d-v*v/a+w*w/a/2)/v+(2*v-w)/a;
        else
            t = (2*sqrt(a*d+w*w/2)-w)/a;
        l = l - d;
        double t2 = (v-w)/a;
        if(l-w*t2-0.5*a*t2*t2>0)
            t2 = t2 + (l-w*t2-0.5*a*t2*t2)/v;
        else
            t2 = (-w+sqrt(w*w+2.0*a*l))/a;
        printf("%.12f
    ",t+t2);
    }
  • 相关阅读:
    把sqlserver2000的备份文件恢复到sqlserver2008中
    application、static、Session、ViewState 的区别
    命令行安装卸载服务(安装卸载.net写的服务)
    C# Windows服务自动安装与注册
    C# System.Guid.NewGuid()
    <base target="_self"/>
    C#写的windows服务,在启动时提示“服务启动后又停止了,一些服务自动停止”
    MVC中validateRequest="false"不起作用
    PL/SQL导出insert语句
    MO和MT
  • 原文地址:https://www.cnblogs.com/qscqesze/p/5313529.html
Copyright © 2011-2022 走看看