Turn the corner (三分)

Turn the corner

Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 151 Accepted Submission(s): 61

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

 

汽车拐弯问题，给定X, Y, l, d判断是否能够拐弯。

我们发现随着角度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。

这里的Calc函数需要比较繁琐的推倒公式：
s = l * cos(θ) + w * sin(θ) - x;
h = s * tan(θ) + w * cos(θ);
其中s为汽车最右边的点离拐角的水平距离, h为里拐点最高的距离, θ范围从0到90。

3分搜索法

 代码：

#include<stdio.h>
#include<string.h>
#include<math.h>
const double pi=2*asin(1.0);
double x,y,l,d;
double geth(double an){
    double s,h;
    s=l*cos(an)-x+d*sin(an);//应该是减x因为车头要对住墙，然后看车尾最高是否大于y 
    h=s*tan(an)+d*cos(an);
    return h;
}
double ABS(double a){
    return a>=0?a:-a;
}
void erfen(){
    double l=0,m,mm,r=pi/2;
//    printf("%lf
",pi);
    while(ABS(r-l)>1e-10){
        m=(l+r)/2;
        mm=(m+r)/2;
        if(geth(m)>=geth(mm))r=mm;
        else l=m;
    }
//    printf("%lf
",geth(m));
    if(geth(l)>y)puts("no");
    else puts("yes");
}
int main(){
    while(~scanf("%lf%lf%lf%lf",&x,&y,&l,&d)){
        erfen();
    }
    return 0;
}