HDU2438 Turn the corner【三分法】【数学几何】
Problem Description
Mr. West bought a new car! So he is travelling around the city.
One day he comes to a vertical corner. The street he is currently in has a width x, the street he wants to turn to has a width y. The car has a length l and a width d.
Can Mr. West go across the corner?
Input
Every line has four real numbers, x, y, l and w.
Proceed to the end of file.
Output
If he can go across the corner, print "yes". Print "no" otherwise.
Sample Input
10 6 13.5 4
10 6 14.5 4
Sample Output
yes
no
题目大意:有一个直角拐角,给你水平道路宽度Y和竖直高度X,再给你汽车的长l,宽w
问:汽车能否通过这个拐角。
思路:假设汽车的宽度大于水平道路宽度Y或是竖直高度X。不管怎样都通只是。接下来
考虑普通情况。
如图:若汽车最左边与墙一直靠紧,则仅仅须要推断右边最高点是否超过了Y。
设θ为汽车与水平方向的夹角,s为汽车最右边的角到拐点的水平距离。那么
s = l*cos(θ) + w*sin(θ)
- x,从而得出 h = s*tan(θ)+w*cos(θ)。
θ角从0~π/2,变化,h则从低到高再究竟,且是一个凸形函数。利用三分方法
得到最高点的h。与Y比較推断能否通过。
#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<cmath>
using namespace std;
const double PI = acos(-1.0);
double x,y,l,w;
double calc(double angle)
{
double s = l*cos(angle) + w*sin(angle) - x;
double h = s*tan(angle) + w*cos(angle);
return h;
}
int main()
{ while(cin >> x >> y >> l >> w)
{
double left,right,mid,midmid;
left = 0;
right = PI/2;
while(right-left >= 1e-7)
{
mid = (left+right)/2;
midmid = (mid+right)/2;
if(calc(mid) > calc(midmid))
right = midmid;
else
left = mid;
}
if(x<w || y<w || calc(mid) > y)
cout << "no" << endl;
else
cout << "yes" << endl;
} return 0;
}
HDU2438 Turn the corner【三分法】【数学几何】的更多相关文章
- 【三分法】hdu2438 Turn the corner
Problem Description Mr. West bought a new car! So he is travelling around the city.One day he comes ...
- HDU2438:Turn the corner(三分)
传送门 分析 根据这张图,我们只要使得h<=y即可,可以发现h是一个凸函数,故使用三分,具体见代码 代码 #include<cstdio> #include<cstring&g ...
- hdu 2348 Turn the corner(三分&&几何)(中等)
Turn the corner Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) ...
- Turn the corner (三分)
Turn the corner Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Tot ...
- hdu 1577 WisKey的眼神 (数学几何)
WisKey的眼神 Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total S ...
- Turn the corner
Problem Description Mr. West bought a new car! So he is travelling around the city. One day he comes ...
- Turn the corner
Turn the corner Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Tot ...
- HDU 2438 Turn the corner(三分查找)
托一个学弟的福,学了一下他的最简便三分写法,然后找了一道三分的题验证了下,AC了一题,写法确实方便,还是我太弱了,漫漫AC路!各路大神,以后你们有啥好的简便写法可以在博客下方留个言或私信我,谢谢了! ...
- hdu 1115 Lifting the Stone (数学几何)
Lifting the Stone Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others ...
随机推荐
- Android开发系列(十六):【Android小游戏成语连连看】第二篇
写的晚了,在分工个Z市高中的一个成绩查询的系统,原系统居然是用VB写的,我不得不佩服原本写系统的那位哥们真能耐得住. 明天搭建下SVN就等着先发project款然后開始项目了.想想有工资进账,心里也为 ...
- julia/pyplot 绘图加入标签和标题
julia 调用matplotlib.pyplot 须要先using pycall 先安装pycall Pkg.add("PyCall") 然后吧. . . 上代码把:(应该是通俗 ...
- Ajax的跨域问题分析
一.Ajax的跨域问题 Ajax是利用javascript内置XMLHttpRequest对象来进行传输的,所以它依赖于XMLHttpRequest对象,而XMLHttpRequest对象却有很多的限 ...
- mvc表单Form提交 --实体
1.方式1:字段加验证 @model MvcWeb.Models.UserInfo @{ ViewBag.Title = "Add"; } <h2>Add</h2 ...
- 关于html(meta的常用的用法)
http://www.haorooms.com/post/html_meta_ds
- Vue中Mixins使用
mixins是一种分发Vue组件中可复用功能的一种灵活方式. mixins是一个JavaScript对象,可以包含组件中的任意选项,比如Vue实例中生命周期的各个钩子函数,也可以是data.compo ...
- atime&&mtime&&ctime区别
- ip---查看网络信息
Linux的ip命令和ifconfig类似,但前者功能更强大,并旨在取代后者. ifconfig属于net-tools.ip属于iproute2 设置一个IP地址,可以使用下列ip命令: ip add ...
- Input/output subsystem having an integrated advanced programmable interrupt controller for use in a personal computer
A computer system is described having one or more host processors, a host chipset and an input/outpu ...
- wpf--------------datagrid全选反选 多选进行删除操作 前后台
前台绑定 <DataGrid.Columns> <DataGridTemplateColumn > <DataGridTemplateColumn.HeaderTempl ...