Codeforces Beta Round #1 C. Ancient Berland Circus

最新推荐文章于 2022-07-13 02:01:45 发布

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最新推荐文章于 2022-07-13 02:01:45 发布

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本篇介绍了一个数学问题的解决方案：已知三个坐标点，如何确定这些点所构成的最小正多边形面积。文章通过示例代码详细解释了计算过程，并考虑了不同角度组合的可能性。

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Nowadays all circuses in Berland have a round arena with diameter 13 meters, but in the past things were different.

In Ancient Berland arenas in circuses were shaped as a regular (equiangular) polygon, the size and the number of angles could vary from one circus to another. In each corner of the arena there was a special pillar, and the rope strung between the pillars marked the arena edges.

Recently the scientists from Berland have discovered the remains of the ancient circus arena. They found only three pillars, the others were destroyed by the time.

You are given the coordinates of these three pillars. Find out what is the smallest area that the arena could have.

Input

The input file consists of three lines, each of them contains a pair of numbers –– coordinates of the pillar. Any coordinate doesn't exceed 1000 by absolute value, and is given with at most six digits after decimal point.

Output

Output the smallest possible area of the ancient arena. This number should be accurate to at least 6 digits after the decimal point. It's guaranteed that the number of angles in the optimal polygon is not larger than 100.

Examples

input
0.000000 0.000000
1.000000 1.000000
0.000000 1.000000

output
1.00000000

菜鸟能力有限，借鉴了大神的博客：http://blog.youkuaiyun.com/shiyuankongbu/article/details/8434560

题意：已知三个点，求这三个点围成的正多边形面积最小。

思路：无，先存在博客里面，以后再来看看！

大神的代码：

#include <cstdio>  
#include <iostream>  
#include <cmath>  
#include <algorithm>  
using namespace std;  
#define pi 3.1415926  
#define eps 10e-3  
typedef struct point  
{  
    double x,y;  
};  
double area(double x0,double y0,double x1,double y1,double x2,double y2)  
{  
    return 0.5*fabs(x0*y1+x2*y0+x1*y2-x2*y1-x0*y2-x1*y0);  
}  
double dis(point a,point b)  
{  
    return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));  
}  
point p[3];  
bool is_ok(int n,double jiao)  
{  
    double cnt=n*jiao/pi;  
    double pnt=floor(cnt+eps);  
    if(cnt-pnt<eps)  
        return true;  
    return false;  
}  
int main()  
{  
    for(int i=0;i<3;i++)  
        scanf("%lf%lf",&p[i].x,&p[i].y);  
    double a=dis(p[0],p[1]);  
    double b=dis(p[0],p[2]);  
    double c=dis(p[1],p[2]);  
    double A=acos((b*b+c*c-a*a)/2/b/c);  
    double B=acos((a*a+c*c-b*b)/2/a/c);  
    double C=acos((b*b+a*a-c*c)/2/b/a);  
    double r=a*b*c/area(p[0].x,p[0].y,p[1].x,p[1].y,p[2].x,p[2].y)/4;  
    //double jiao=max(A,max(B,C)); 开始我觉得下面的is_ok函数只要判断jiao就可以了  
    // 其实还是会WA 第五组数据。。。WA的时候答案是输出的两倍，Wa的厉害了  
    int n;  
    for(n=3;n<=100;n++)  
        if(is_ok(n,A)&&is_ok(n,B)&&is_ok(n,C))  
            break;  
    double tag=2*pi/n;  
    double Area=n*r*r*sin(tag)/2;  
    printf("%.8lf\n",Area);  
    return 0;  
}