lazy gege

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

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

题意：

给你一张边长为L的正方形桌子，一本A*B的笔记本，求笔记本放桌子上的最小面积。

题解：只需要吧笔记本的重心放在桌子上就行了

CODE：

#include<cstdio>
#include<algorithm>
#include<iostream>
#include<cstring>
#include<cmath>

using namespace std;

double a,b,c;

int main() {
    int t;
    cin>>t;
    while(t--) {
        scanf("%lf%lf%lf",&a,&b,&c);
        if(b<c)
            swap(b,c);
        double x=b/2;
        double y=c/2;
        double r=sqrt(2*a*a);
        double ans;
        if(r<y) {
            ans=a*a;
        } else {
            if(y<r/2) {
                ans=y*y;
            } else {
                double z=r-y;
                ans=a*a-z*z;
            }
        }
        printf("%.4f\n",ans);
    }
    return 0;
}