题意:在一个拐弯的街道,一面宽x,一面宽y,车宽为d,车长为l为这辆汽车能否拐过这个弯
思路:
通过模拟得出h是先增大后减小的,一个凸函数,果断三分求极值,方程,s是最右面的点到右数第一条竖线的水平距离,h是图上的高
方程
s = sin(a)*w+l*cos(a) - x;
h = s * tan(a) + cos(a) * w;
h是一个凸函数,可以用三分求出当h取最大时的角度a,再把a带入方程得到h1,然后和h比较 如果h1 > h则过不去,反之能过去
#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
using namespace std;
double x,y,w,l;
const double PI = acos(-1.0);
const double eps = 1e-9;
double f(double a)
{
double sum = sin(a)*w + l*cos(a) - x;
return sum*tan(a)+cos(a)*w;
}
double f1(double a)
{
return l*cos(a) - x+2*sin(a)*w;
}
double sanfen(double a)
{
double left = 0,right = a,mid,midmid;
while(right - left > eps)
{
mid = (left + right) / 2;
midmid = ( mid + right) / 2;
if(f(mid) < f(midmid))
{
left = mid;
}
else
{
right = midmid;
}
}
return (mid + midmid) / 2;
}
int main()
{
while(scanf("%lf%lf%lf%lf",&x,&y,&l,&w) != EOF)
{
double h = sanfen(PI/2);
if(f(h) > y)
printf("no\n");
else
printf("yes\n");
}
return 0;
}