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CodeForces - 630P Area of a Star （数学几何&正弦定理求面积）

2016-03-10 12:18 337 查看
CodeForces
- 630P

Area of a Star

Time Limit: 500MS Memory Limit: 65536KB 64bit IO Format: %I64d & %I64u
Submit Status

Description

It was decided in IT City to distinguish successes of local IT companies by awards in the form of stars covered with gold from one side. To order the stars it is necessary to estimate order cost that depends on the area of gold-plating. Write a program that
can calculate the area of a star.

A "star" figure having n ≥ 5 corners where n is a prime number is constructed
the following way. On the circle of radius rn points are selected so that the distances between the
adjacent ones are equal. Then every point is connected by a segment with two maximally distant points. All areas bounded by the segments parts are the figure parts.



Input

The only line of the input contains two integers n (5 ≤ n < 109, n is
prime) and r (1 ≤ r ≤ 109) — the number of the star corners and the radius
of the circumcircle correspondingly.

Output

Output one number — the star area. The relative error of your answer should not be greater than 10 - 7.

Sample Input

Input
7 10


Output
108.395919545675


Source

Experimental Educational Round: VolBIT Formulas Blitz

//题意：

给出n,r表示在半径为r的圆里有一个n角星，问这个n角星面积为多少。

//思路：

每个角顶点到圆心可以分为两个全等三角形，先利用圆周角和圆心角的关系求出三角形的三个角，再利用正弦定理，可以求出三角形的边长，再利用s=(1/2)*a*b*sin(c),求出每个三角形的面积，再将s*2*n(2*n为全等三角形的个数)即为所求。

#include<stdio.h>
#include<string.h>
#include<math.h>
#include<algorithm>
#define INF 0x3f3f3f3f
#define ll long long
#define N 10010
#define M 1000000007
#define PI acos(-1.0)
using namespace std;
int main()
{
double n,r;
while(scanf("%lf%lf",&n,&r)!=EOF)
{
double x1=PI/n;
double x2=PI/(2*n);
double x3=PI-x1-x2;
double len=(r*sin(x2))/sin(x3);
double s=0.5*len*r*sin(x1);
s=s*2*n;
printf("%.8lf\n",s);
}
return 0;
}


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