Time Limit: 2000/1000 MS (Java/Others)   

 Memory Limit: 65536/65536 K (Java/Others)

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 題意：  求變成為length的正方形中陰影部分的面積; 思路：

建座標系積分就好了;以參考資料中的D點為原點,然後得到兩個圓的方程,聯立求交點得到積分割槽間,然後積分,積分的時候借用三角函式;  http://zhidao.baidu.com/question/571519797

另一種解法也可以用以正方形的中心為原點,以對角線為水平軸,然後得到兩個圓的方程,聯立求交點得到積分割槽間,然後積分。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <queue>
#include <map>
#include <set>
#include <stack>
#include <vector>
#include <list>
#define LL long long
#define eps 1e-8
#define maxn 110
#define mod 100000007
#define inf 0x3f3f3f3f
#define mid(a,b) ((a+b)>>1)
#define IN freopen("in.txt","r",stdin);
using namespace std;

int main(int argc, char const *argv[])
{
    //IN;

    int t; cin >> t;
    while(t--)
    {
        double a; scanf("%lf", &a);

        //double ans = a*a/4.0 * asin(sqrt(14.0)/4.0);
        //ans -= a*a * asin(sqrt(14.0)/8.0);
        //ans += a*a * sqrt(7.0) / 8.0;
        //ans *= 2.0;

        double ans = atan(sqrt(7.0)) / 4.0;
        ans -= atan(sqrt(7.0)/5.0);
        ans += sqrt(7.0) / 8.0;
        ans *= a*a*2.0;

        printf("%.2f\n", ans);
    }

    return 0;
}