HDU——2438 Turn the corner

客官°小女子只卖身不卖艺 2022-03-25 02:39 167阅读 0赞

# Turn the corner #

**Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)  
Total Submission(s): 3952    Accepted Submission(s): 1658**

**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?  
![2438-1.jpg][]

**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

**Source**

[2008 Asia Harbin Regional Contest Online][]

**Recommend**

gaojie   |   We have carefully selected several similar problems for you:  [2441][] [2442][] [2443][] [2440][] [2439][]

题目大意：有一个带拐角的街道，水平宽度为x，竖直宽度为y 有一辆长为l宽为w的小汽车从水平街道驶来，问你它能否进入竖直街道

思路：以汽车的上端边所在直线进行分析，，将右下角那个点为坐标原点建立直角坐标系 求得直线Y=tan@\*x+L\*sin@+D/cos@;此时Y=X，求解x，再判断x与-y的关系，三分@，（@表示角，这里三分的是角，不是边或其他）

图为：

![watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L2xlZTM3MTA0Mg_size_16_color_FFFFFF_t_70][]

代码：

/*
     题意：有一个带拐角的街道，水平宽度为x，竖直宽度为y
     有一辆长为l宽为w的小汽车从水平街道驶来，问你它能否进入竖直街道
     思路：以汽车的上端边所在直线进行分析，，将右下角那个点为坐标原点建立直角坐标系
     求得直线Y=tan@*x+L*sin@+D/cos@;此时Y=X，求解x，再判断x与-y的关系，三分@
    */
    #include<set>
    #include<map>
    #include<stack>
    #include<queue>
    #include<string>
    #include<cmath>
    #include<cstdio>
    #include<cstring>
    #include<cstdlib>
    #include<iostream>
    #include<algorithm>
    using namespace std;
    #define eps 1e-7
    #define ll long long
    #define inf 0x3f3f3f3f
    #define bug printf("mmp\n");
    #define ul unsigned long long
    #define mm(a,b) memset(a,b,sizeof(a))
    #define T()int test,q=1,tt;scanf("%d",&test),tt=test;while(test--)
    const int N=1e2+10;
    const int maxn=1e6+100;
    const int mod=1000000007;
    const double pi=acos(-1.0);
    double x,y,l,w;
    double finds(double du)
    {
        return (-x+l*sin(du)+w/cos(du))/tan(du);
    }
    double three_search(double low ,double high)
    {
        double mid=0,mids=0;
        while(high-low>eps)
        {
            mid=(low+high)*0.5;
            mids=(mid+high)*0.5;
            if(finds(mid)>finds(mids))
               high=mids;
            else
                low=mid;
    
        }
        return finds(mid);
    }
    int main()
    {
        while(scanf("%lf%lf%lf%lf",&x,&y,&l,&w)!=EOF)
        {
            if(x<w||y<w)
            {
                printf("no\n");
                continue;
            }
            if(three_search(0,pi/2.0)<=y)
                printf("yes\n");
            else
                printf("no\n");
    
        }
        return 0;
    }

[2438-1.jpg]: http://acm.hdu.edu.cn/data/images/2438-1.jpg
[2008 Asia Harbin Regional Contest Online]: http://acm.hdu.edu.cn/search.php?field=problem&key=2008+Asia+Harbin+Regional+Contest+Online&source=1&searchmode=source
[2441]: http://acm.hdu.edu.cn/showproblem.php?pid=2441
[2442]: http://acm.hdu.edu.cn/showproblem.php?pid=2442
[2443]: http://acm.hdu.edu.cn/showproblem.php?pid=2443
[2440]: http://acm.hdu.edu.cn/showproblem.php?pid=2440
[2439]: http://acm.hdu.edu.cn/showproblem.php?pid=2439
[watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L2xlZTM3MTA0Mg_size_16_color_FFFFFF_t_70]: /images/20220325/6c72482e8b0e45a683b487cc02e70bde.png