ZOJ 3203 Light Bulb（三分）

Light Bulb

---

Time Limit:1 Second     Memory Limit:32768 KB

---

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 integerT (T <= 100), indicating the number of cases.

Each test case contains three real numbersH,h andD in one line.H is the height of the light bulb whileh 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, andH -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

3
2 1 0.5
2 0.5 3
4 3 4

 

Sample Output

1.000

0.750
4.000

题意：如图，给出的条件，求L的最大值。

分析：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<stdio.h>
#include<string.h>
#include<math.h>
#include<iostream>
#include<algorithm>
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<temp2)printf("%.3lf\n",h*D/H);
        else
        {
            double ans=D+H-temp-(H-h)*D/temp;
            printf("%.3lf\n",ans);
        }
    }
    return 0;
}

 

 

三分法：

#include<stdio.h>
#include<iostream>
#include<algorithm>
#include<string.h>
#include<math.h>
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;
}