Convex

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1378 Accepted Submission(s): 923

Problem Description 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

Input 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 contain N integers indicating the angles. The sum of the N numbers is always 360.

Output 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.

Sample Input 4 1 90 90 90 90 6 1 60 60 60 60 60 60 Sample Output 2.000 2.598

題意：

給n個點，每個點距頂點的距離都是d，給出每兩點與頂點連線之間的角度，求所形成的凸包的上表面積。

三角形S==0.5*a*b*sinc

註意 ，sin操作的是弧度而非角度，故要轉換成弧度：

弧度==角度*PI/180 (PI=acos(-1.0））。

AC代碼：

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 
 4 #define PI acos(-1.0) 
 5 
 6 int main(){
 7     ios::sync_with_stdio(false);
 8     int n,d,x;
 9     double sum=0;
10     while(cin>>n>>d){
11         sum=0;
12         for(int i=0;i<n;i++){
13             cin>>x;
14             sum+=0.5*d*d*sin(x*1.0*PI/180);
15         }
16         printf("%.3lf\n",sum);
17     }
18     return 0;
19 }

HDU-5979