ACDream 1067 Convex

Convex
Time Limit: 2000/1000MS (Java/Others) Memory Limit: 128000/64000KB (Java/Others)
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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, D (3 ≤ N ≤ 10, 1 ≤ D ≤ 10) indicating the number of the points and their distance to origin.
The next lines contains 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
正弦定理，题意告诉我们有一个凸边形，上面任意去n个点，点和相邻的点给你角度，要你算出各个点之间的面积和。

    #include <iostream>#include <cstdio>#include <cmath>#define pi 3.1415926536using namespace std;int main(){    int n,d;    double angle[20];    while(scanf("%d%d",&n,&d)!=EOF)    {        for(int i=0;i<n;i++)        scanf("%lf",&angle[i]);        double s=0;        for(int i=0;i<n;i++)        s+=0.5*d*d*sin((angle[i]/180)*pi);        printf("%.3lf\n",s);    }    return 0;}