I - Convex HDU - 5979

@小阿宝

于 2020-12-09 11:16:27 发布

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本文链接：https://blog.youkuaiyun.com/m0_46062697/article/details/110918788

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该博客介绍了一个特殊的凸多边形问题，其中所有点到原点的距离相等。给定每个三角形的一个内角和边长，任务是计算多边形的面积。通过将多边形划分为多个三角形并应用三角形面积公式，可以求解答案。题目提供了多组测试用例，并要求输出结果保留3位小数。

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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

题解：给出n边形，然后将n边形分成n个三角形，告诉你每个三角形的一个角度，然后告诉你三角形边长。让你求出n边形的面积。

**其实就是求三角形面积 s=absinc/2;**

#include <iostream>
#include <bits/stdc++.h>
using namespace std;
double pi=acos(-1);
int main()
{
    int n,a;
    while(~scanf("%d %d",&n,&a))
    {
        int b[12];
        double s=0;
        for(int i=0; i<n; i++)
        {
            scanf("%d",&b[i]);
            double f=double(b[i])/double(180)*pi;
            s=s+sin(f);
        }
        double sum=double(a)*double(a)*s/2.0;
        printf("%.3lf\n",sum);
    }
    return 0;
}