学步园

We have a special convex that all points have the same distance to origin point.
As you know we can get N segments after linking the origin point and the points on the convex. We can also get N angles between each pair of the neighbor segments.
Now give you the data about the angle, please calculate the area of the convex.

There are multiple test cases.
The first line contains two integer N and D indicating the number of the points and their distance to origin. (3 <= N <= 10, 1 <= D <= 10)
The next lines contains N integers indicating the angles. The sum of the N numbers is always 360.

For each test case output one float numbers indicating the area of the convex. The printed values should have 3 digits after the decimal point.

4 1
90 90 90 90
6 1
60 60 60 60 60 60

2.000
2.598

多边形分成多个三角形，利用面积s=1/2*d*d*sin(角度的弧度)

#include<stdio.h>
#include<math.h>
#define PI 3.1415926
int main()
{
    int n;
    double d,c,sum;
    while(scanf("%d%lf",&n,&d)>0)
    {
        sum=0;
        while(n--)
        {
            scanf("%lf",&c);
            sum+=sin(c/180*PI);
        }
        sum=sum/2*d*d;
        printf("%.3lf\n",sum);
    }
}