HDU 3349 lazy gege（水题）

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=3349

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

 

题目大意：矩形书本a*b放在正方形桌子上L*L，求最小接触面积

题目思路：由正方形顶点对着书本肯定最小，刚好放到矩形宽的一半即可

代码：

#include<cstdio>
#include<cmath>
#include<cstring>
#include<string>
#include<cstdlib>
#include<algorithm>
#include<iostream>
#include<queue>
#include<stack>
#include<map>

using namespace std;

#define FOU(i,x,y) for(int i=x;i<=y;i++)
#define FOD(i,x,y) for(int i=x;i>=y;i--)
#define MEM(a,val) memset(a,val,sizeof(a))
#define PI acos(-1.0)

const double EXP = 1e-9;
typedef long long ll;
typedef unsigned long long ull;
const int INF = 0x3f3f3f3f;
const ll MINF = 0x3f3f3f3f3f3f3f3f;
const int mod = 1e9+7;
const int N = 1e6+5;

int main()
{
    //freopen("in.txt","r",stdin);
    //freopen("out.txt","w",stdout);
    std::ios::sync_with_stdio(false);
    int t;
    double L,a,b;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%lf%lf%lf",&L,&a,&b);
        if(a>b)
            swap(a,b);
        double len = sqrt(2.0)*L/2;  //对角线一半
        double tmp = a/2;
        double x=2*len-tmp;
        if(tmp>len*2)
            printf("%.4lf\n",L*L);
        else if(tmp<len)
            printf("%.4lf\n",tmp*tmp);
        else
            printf("%.4lf\n",L*L-x*x);
    }
    return 0;
}