2016ICPC-大连 Convex （几何）

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
InputThere 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 contain N integers indicating the angles. The sum of the N numbers is always 360.
OutputFor each test case output one float numbers indicating the area of the convex. The printed values should have 3 digits after the decimal point.
Sample Input

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

Sample Output

2.000
2.598

S=1/2*a*b*sinA
把各个小三角形的面积加起来就好
水题

 1 #include <iostream>
 2 #include<cstdio>
 3 #include<algorithm>
 4 #include<cstring>
 5 #include<string>
 6 #include<queue>
 7 #include<vector>
 8 #include<cmath>
 9 using namespace std;
10 const double pi=acos(-1.0);
11 int n,d,x;
12 int main()
13 {
14      while(~scanf("%d%d",&n,&d))
15     {
16         double sum=0;
17        for(int i=1;i<=n;i++)
18        {
19            scanf("%d",&x);
20            sum+=0.5*d*d*sin(x*1.0*pi/180.0);
21        }
22        printf("%.3lf\n",sum);
23     }
24     return 0;
25 }

2016ICPC-大连 Convex （几何）

原文：http://www.cnblogs.com/Annetree/p/7612447.html