Turn the corner
Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1930 Accepted Submission(s): 736
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;
}