解題思路：

看懂這個圖，這個題就能A掉了。
下面是AC程式碼：

#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
using namespace std;
#define PI acos(-1.0)

int main()
{
    int r1,r,t;
    scanf("%d",&t);
    while(t--)
    {scanf("%d",&r1);
        r=(double)r1;
        double
 
 x=1.25/(sqrt(2.0));
        double jiaodu=(acos(x))*2.0;
        double hudu=acos(x)*2.0;
        double area1=1.0*r*r*hudu*0.5;
        double area2=0.5*r*r*sin(hudu);
        double area=area1-area2;
        double hudu2=2.0*PI-2.0*acos(-sqrt(2.0)/4.0);
        double area3=0.25*r*r*hudu2/2.0;
        area=area3-area-0.25
 
*r*r*sin(hudu2)*0.5;
        area*=2.0;
        printf("%.2lf\n",area);
    }
    return 0;
}