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| polygon | ||||||
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| Description | ||||||
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We have a special polygon that all points have the same distance to original point.As you know we can get N segments after linking the original point and the points on the polygon, then 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 polygon. |
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| Input | ||||||
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There are multiple test cases. The first line contains two integer N and D indicating the number of the points and their distance to original point. (3 <= N <= 10, 1 <= D <= 10) The next lines contains N integers indicating the angles. The sum of the N numbers is always 360.
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| Output | ||||||
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For each test case output one float numbe indicating the area of the polygon. The printed value should have 3 digits after the decimal point.
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| Sample Input | ||||||
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4 1 90 90 90 90 6 1 60 60 60 60 60 60 |
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| Sample Output | ||||||
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2.000 2.598 |
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| Hint | ||||||
| 原点在多边形内部，各顶点与原点相连得到相邻两线段夹角小于180度 | ||||||
| Author | ||||||
| 陈禹@HRBUST |
题目大意： 给你一个多边形，多边形中间有一个点，多边形上的点距离这个原点都相等，已知这个距离和相邻两个点与这个原点组成的夹角
求面积
#include <math.h> #include <stdio.h> #include <string.h> #include <algorithm> #include <iostream> using namespace std; int N; double d; double PI=atan(1)*8; int main() { //cout<<PI<<endl; while(~scanf("%d%lf",&N,&d)) { double ans; double sum=0; int i; for(i=0;i<N;i++) { scanf("%lf",&ans); ans=ans/360*PI; sum+=d*d*sin(ans)/2; // cout<<sin(ans)<<endl; } printf("%.3lf\n",sum); } return 0; }
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原文地址：http://www.cnblogs.com/beige1315402725/p/4921812.html
