Hard problem
Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 843    Accepted Submission(s): 527
Problem Description
cjj is fun with math problem. One day he found a Olympic Mathematics problem for primary school students. It is too difficult for cjj. Can you solve it?


Give you the side length of the square L, you need to calculate the shaded area in the picture.


The full circle is the inscribed circle of the square, and the center of two quarter circle is the vertex of square, and its radius is the length of the square.
 
Input

The first line contains a integer T(1<=T<=10000), means the number of the test case. Each case contains one line with integer l(1<=l<=10000).

Output
For each test case, print one line, the shade area in the picture. The answer is round to two digit.

Sample Input1 1
Sample Output0.29

（area_S表示S的面積）

以正方形左下角為座標軸原點，分別以正方形下邊、左邊為X軸，Y軸，以左上角為圓心（0,l）半徑為l作圓S1，以右下角（l,0）為圓心半徑為l作圓S2，正方形內切圓是以(l/2,l/2)為圓心，l/2為半徑的圓S。

思路:求出圓S與圓S1相交部分面積area1和圓S與圓S2的相交面積area2,設S中空白部分面積為X，有area1+area2-X=area_S;求出X，用area_S-X即得陰影部分面積。area1與area2是相等的，所以用求出一個即可。

只要敲出兩圓相交面積模板就行了：

#include<stdio.h>
#include<iostream>
#include<cmath>
using namespace std;

#define pi acos(-1.0)

typedef struct node
{
	double x;
	double y;
}point;
//兩圓相交求陰影面積
double AREA(point a, double r1, point b, double r2)
{
	double d = sqrt((a.x-b.x)*(a.x-b.x) + (a.y-b.y)*(a.y-b.y));
	if (d >= r1+r2)
		return 0;
	if (r1>r2)
	{
		double tmp = r1;
		r1 = r2;
		r2 = tmp;
	}
	if(r2 - r1 >= d)
		return pi*r1*r1;
	double ang1=acos((r1*r1+d*d-r2*r2)/(2*r1*d));
	double ang2=acos((r2*r2+d*d-r1*r1)/(2*r2*d));
	return ang1*r1*r1 + ang2*r2*r2 - r1*d*sin(ang1);
}

int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        int n;
        scanf("%d",&n);
        point a, b;
        a.x=n*1.0/2, a.y=n*1.0/2;
	    b.x=n*1.0, b.y=0.0;
	   double result = AREA(a, n*1.0/2.0, b, n*1.0);
	   printf("%.2lf\n",2.0*pi*(n*1.0/2.0)*(n*1.0/2.0)-2.0*result);
    }
	return 0;
}