Codeforces 1C Ancient Berland Circus

最新推荐文章于 2021-06-28 17:54:16 发布

原创最新推荐文章于 2021-06-28 17:54:16 发布 · 1.6k 阅读

1 ·

CC 4.0 BY-SA版权

本文探讨如何通过给定三个支柱坐标，计算古代罗马斗兽场可能形成的最小正多边形面积。利用海伦公式、正弦定理和余弦定理解决问题。

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.

Sample test(s)

input
0.000000 0.000000
1.000000 1.000000
0.000000 1.000000

output
1.00000000

这是我第一题几何题。纪念下。

题意：有一个圆形的斗兽场，圆上有X个树桩，刚好能组成一个正多边形的舞台。但是，有3个树桩不见了。

现在给你这3个树桩的坐标，让你判断这个舞台最小的面积。（数组绝对合法。绝对能组成三角形）不会大于100边型。

思路：

#include <stdio.h>
#include <math.h>
#define pi acos(-1.0)
#include <cmath>
#define feq(a, b) (fabs((a)-(b))<1E-2)
using namespace std; 
double fgcd(double a, double b){
     if (feq(a, 0))
         return b;
     if (feq(b, 0))
         return a;
     return fgcd(b, fmod(a, b));
}
double diameter(double a, double b, double c) {       
   	double p = (a + b + c) / 2;
   	double s = sqrt(p * (p - a) * (p - b) * (p - c));
   	return a * b * c / (4 *s);
} 
int main() {
	double x[3], y[3], d[3];
	int i =  0;
	while(~scanf("%lf%lf", &x[i], &y[i])) {
		i++;
		if(i == 3) {
			for(int j = 0; j < 3; ++j)
				d[j] = sqrt((x[j] - x[(j+1)%3]) * (x[j] - x[(j+1)%3]) + (y[j] - y[(j+1)%3]) * (y[j] - y[(j+1)%3]));
			double R = diameter(d[0], d[1], d[2]);
			double angle[4];
			for(int j = 0; j < 2; ++j)
			 	angle[j] = acos(1 - d[j] * d[j] / (2 * R * R));
	 		angle[2] = 2 * pi - angle[0] - angle[1];
 			angle[3] = fgcd(angle[0], fgcd(angle[1], angle[2]));
			printf("%.6lf\n", R * R * sin(angle[3]) / 2 * (2 * pi / angle[3]));
			i = 0;
		}
	}
}

如图给出A(x0, y0) B(x1, y1) C(x2, y2)

1.求3边a,b,c

2. 先求外接圆半径。（一定存在）

海伦公式 + 正弦定理得 R = a * b * c / (4 * S) S = sqrt(q * (q - a) * (q - b) * (q -c)); q = (a + b + c) / 2;

-----因为是正多边形。那么只要求出一边与两半径围成的面积 * N 就好。

3. 余弦定理求3个角。求最大公约数就是正多边形每一份最小的角度。

4. 最后就是用正弦求面积了。 S = R * R * sin( angle ) / 2 * (倍数) （竟然坐标是合法的，倍数 = 2 * pi / angle);