HDU-5979

Posted 2020-10-15 Kiven#5197

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

 

题意：

给n个点，每个点距顶点的距离都是d，给出每两点与顶点连线之间的角度，求所形成的凸包的上表面积。

 

三角形S==0.5*a*b*sinc

 

注意 ，sin操作的是弧度而非角度，故要转换成弧度：

 

弧度==角度*PI/180   (PI=acos(-1.0））。

 

 

AC代码：

 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 }