zoj 3203 Light Bulb(公式推导|三分法)(简单)

Light Bulb

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

3
2 1 0.5
2 0.5 3
4 3 4

Sample Output

1.000
0.750

4.000

公式推导：
#include<stdio.h>
#include<string.h>
#include<math.h>
#include<iostream>
#include<algorithm>
using namespace std;

int main()
{
    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<iostream>
#include<cstdio>
using namespace std;
const double eps=1e-9;
double H,h,D;

double getL(double x)
{
    return D-x+H-(H-h)*D/x;
}

void search()
{
    double mid,midmid;
    double l=((H-h)*D)/H,r=D;
    while(r-l>eps)
    {
        mid=(l+r)/2;
        midmid=(mid+r)/2;
        if(getL(mid)<getL(midmid))
            l=mid;
        else
            r=midmid;
    }
    printf("%.3lf\n",getL(l));
}

int main()
{
    int t;
    cin>>t;
    while(t--)
    {
        cin>>H>>h>>D;
        search();
    }
}

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