### HDU - 5858（acos返回的是弧度）

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 Input

1 1

Sample Output

0.29

```#include<stdio.h>
#include<string.h>
#include<math.h>
#include<queue>
#include<algorithm>
#define inf 0x3f3f3f3f
#define ll long long
#define maxx 5000000
using namespace std;
#define pi acos(-1)
int main()
{
int t;
double l;
scanf("%d",&t);
while(t--)
{
scanf("%lf",&l);
double r=l/2.0;
double x=sqrt(2.0*l*l)/2.0;
double a = (x*x + l*l -r*r)/(2.0*x*l);
double s1 = l*x*sin(acos(a));
double s2 = pi*l*l*2.0*acos(a)*180.0/pi/360.0;
double b = (r*r+x*x-l*l)/(2.0*r*x);
b = acos(b)*180.0/pi;
b = 360.0 - 2.0*b;
double s3 = pi*r*r*b/360.0;
double ans = s3-(s2 - s1);
printf("%.2lf\n",ans*2.0);
}
}
``` 