ZOJ Problem Set - 3203

Light Bulb

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Time Limit: 1 Second     

Memory Limit: 32768 KB

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Compared to wildleopard's wealthiness, his brother mildleopard is rather poor. His house is narrow and he has only one light bulb in his house. Every night, he is wandering in his incommodious house, thinking of how to earn more money. One day, he found that the length of his shadow was changing from time to time while walking between the light bulb and the wall of his house. A sudden thought ran through his mind and he wanted to know the maximum length of his shadow.

Input

The first line of the input contains an integer T (T <= 100), indicating the number of cases.

Each test case contains three real numbers H, h and D in one line. H is the height of the light bulb while h is the height of mildleopard. D is distance between the light bulb and the wall. All numbers are in range from 10^-2 to 10³, both inclusive, and H - h >= 10^-2.

Output

For each test case, output the maximum length of mildleopard's shadow in one line, accurate up to three decimal places..

Sample Input

32 1 0.52 0.5 34 3 4

 

Sample Output

1.0000.7504.000

本来此题直接推出了公式，直接找到了最值点的。

但是看到网上好多用三分法做的，就复习了下三分法，因为是凸的，所以用三分。

 

 

公式法：

#include
      
       #include
       
        #include
        
         #include
         
          #include
          
           using namespace std;int main(){   // freopen("in.txt","r",stdin);   // freopen("out.txt","w",stdout);    int T;    double H,h,D;    scanf("%d",&T);    while(T--)    {        scanf("%lf%lf%lf",&H,&h,&D);        double temp=sqrt((H-h)*D);        double temp2=(H-h)*D/H;        if(temp>=D)printf("%.3lf\n",h);        else if(temp
          
         
        
       
      

 

 

三分法：

#include
      
       #include
       
        #include
        
         #include
         
          #include
          
           using namespace std;const double eps=1e-9;double D,H,h;double calc(double x){    return D-x+H-(H-h)*D/x;}double solve(double l,double r){    double mid,midmid;    double d1,d2;    do    {        mid=(l+r)/2;        midmid=(mid+r)/2;        d1=calc(mid);        d2=calc(midmid);        if(d1>=d2) r=midmid;//这里要注意        else l=mid;    }    while(r-l>=eps);    return d1;}int main(){    int T;    scanf("%d",&T);    while(T--)    {        scanf("%lf%lf%lf",&H,&h,&D);        printf("%.3lf\n",solve((H-h)*D/H,D));    }    return 0;}