HDU 3349

lazy gege
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1391 Accepted Submission(s): 586

Problem Description
Gege hasn’t tidied his desk for long,now his desk is full of things.
This morning Gege bought a notebook,while to find somewhise to put it troubles him.
He wants to tidy a small area of the desk, leaving an empty area, and put the notebook there, the notebook shouldn’t fall off the desk when putting there.
The desk is a square and the notebook is a rectangle, area of the desk may be smaller than the notebook.
here’re two possible conditions:

Can you tell Gege the smallest area he must tidy to put his notebook?

Input
T(T<=100) in the first line is the case number.
The next T lines each has 3 real numbers, L,A,B(0< L,A,B <= 1000).
L is the side length of the square desk.
A,B is length and width of the rectangle notebook.

Output
For each case, output a real number with 4 decimal(printf(“%.4lf”,ans) is OK), indicating the smallest area Gege should tidy.

Sample Input

3
10.1 20 10
3.0 20 10
30.5 20.4 19.6

Sample Output

25.0000
9.0000
96.0400

題解：
為了將書放在桌子上並儘可能小的佔用空間，當重心放在上面即可；
沿著桌子的稜角去放書，三種情況：
1：
2：
3：

程式碼：

#include<iostream>
#include<stdio.h>
#include<math.h>
using namespace std;
int main()
{
    int T;
    cin >>T;
    while(T--){
        double L,A,B;
        cin >> L >>A >> B;
        double x;
        x=L/sqrt(2);
        double Min;
        Min=min(A,B);
        double s;
        if(x>=Min/2) {
            s=Min*Min/4;
        }
        else if(x*2<=Min/2){
            s=L*L;
        }
        else {
            s=L*L-(sqrt(2)*L-Min/2)*(sqrt(2)*L-Min/2);
        }
        printf("%.4f\n",s);
    }
    return 0;
}