hdu 5979 Convex（水，求面积）

Convex

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 670    Accepted Submission(s): 438

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

 

Source

2016ACM/ICPC亚洲区大连站-重现赛（感谢大连海事大学）

 

Recommend

 

 

 

求多个三角形的面积和，

三角形面积 = absinC/2，C为a和b的夹角

 

#include<bits/stdc++.h>
using namespace std;

#define PI 3.1415926

int main()
{
    //printf("%f
", sin(90/180.0 * PI));
    int n,len,ang;
    while(~scanf("%d%d",&n,&len))
    {
        double sum=0;
        for(int i=0;i<n;i++){
            scanf("%d",&ang);
            sum+=len*len*sin(ang/180.0 * PI)*0.5;
        }
        printf("%.3f
",sum);
    }
    return 0;
}