山西坦达清洁设备有限公司

现在的位置: 主页 > 联系方式 > 文章列表

文章正文

hdu 2348 Turn the corner(三分几何)(中等)

作者:山西坦达清洁设备有限公司 来源:www.sxtdqj.com 发布时间:2017-09-02 11:21:22
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)
Total Submission(s): 2229 Accepted Submission(s): 856


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

题意:

已知汽车的长和宽,l和w,以及俩条路的宽为x和y,汽车所处道路宽为x ,问汽车能否顺利转弯?

分析:汽车能否顺利转弯取决于在极限情况下,随着角度的变化,汽车离对面路的距离是否大于等于0

如图中

\


在上图中需要计算转弯过程中h 的最大值是否小于等于y很明显,随着角度θ的增大,最大高度h先增长后减小,即为凸性函数,可以用三分法来求解

代码:

#include #include #include #include using namespace std; #define pi 3.141592653 double x,y,l,w; double cal(double a) { double s=l*cos(a)+w*sin(a)-x; double h=s*tan(a)+w*cos(a); return h; } int main() { while(scanf(%lf %lf %lf %lf,&x,&y,&l,&w)!=EOF) { double left=0.0,right=pi/2; double lm,rm; while(fabs(right-left)>1e-6) { lm=(left*2.0+right)/3.0; rm=(left+right*2.0)/3.0; if(cal(lm)>cal(rm)) right=rm; else left=lm; } if(cal(left)<=y) printf(yes ); else printf(no ); } return 0; }

企业建站2800元起,携手武汉肥猫科技,做一个有见地的颜值派!更多优惠请戳:武汉网页设计 http://www.feimao666.com


COPYRIGHT © 2015 山西坦达清洁设备有限公司 ALL RIGHTS RESERVED.
网站地图 技术支持:肥猫科技
精彩专题:网站建设
购买本站友情链接、项目合作请联系客服QQ:2500-38-100