Ancient Berland Circus
2016-05-02 21:08
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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.
Sample Input
Input
0.000000 0.000000
1.000000 1.000000
0.000000 1.000000
Output
1.00000000
内容题目大意是给出正多边形上的三个点坐标, 求这个正多边形的面积.
由于是确定是正多边形,所以一定存在外接圆.所以可以分为如下几步:
海伦公式: p=(a+b+c)/2 S=√p(p-a)(p-b)(p-c)
1.求外接圆半径r=abc/4S
2.由余弦定理求出三个圆心角ang[3]
(要注意的是,有可能有三个点在同一段半圆弧上,这是第三个圆心角应该用2π-ang[0]-ang[1], 所以干脆全部都是ang[2]=2π-ang[0]-ang[1])
3.求这三个角的最大公约数为A, 那么这就是一个正2π/A边形.
4.一个小三角形的面积S=1/2·r * r * sinA
5.nS即为所求.
感觉很多东西不了解,我是看了别人的代码才懂的
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 Input
Input
0.000000 0.000000
1.000000 1.000000
0.000000 1.000000
Output
1.00000000
内容题目大意是给出正多边形上的三个点坐标, 求这个正多边形的面积.
由于是确定是正多边形,所以一定存在外接圆.所以可以分为如下几步:
海伦公式: p=(a+b+c)/2 S=√p(p-a)(p-b)(p-c)
1.求外接圆半径r=abc/4S
2.由余弦定理求出三个圆心角ang[3]
(要注意的是,有可能有三个点在同一段半圆弧上,这是第三个圆心角应该用2π-ang[0]-ang[1], 所以干脆全部都是ang[2]=2π-ang[0]-ang[1])
3.求这三个角的最大公约数为A, 那么这就是一个正2π/A边形.
4.一个小三角形的面积S=1/2·r * r * sinA
5.nS即为所求.
感觉很多东西不了解,我是看了别人的代码才懂的
这里写代码片 #include <cstdio> #include <cstring> #include <math.h> using namespace std ; #define PI acos(-1) #define eqs 0.01 double gcd(double a,double b) { return a < eqs ? b : gcd(fmod(b,a),a);//fmod() 用来对浮点数进行取模(求余) //头文件为#include <math.h>fmod()函数可以对浮点型数据进行取模运算,后一个数可为0,这时函数返回NaN。 //计算后结果的符号与前者(x)相同,如果前者是较小的数,后者是较大的数,那么结果直接为较小的数。 } int main() { double x1 , y1 , x2 , y2 , x3 , y3 ; double a , b , c , p , s , r , k ; double A , B , C ; scanf("%lf %lf %lf %lf %lf %lf", &x1, &y1, &x2, &y2, &x3, &y3); a = sqrt( (x1-x2)*(x1-x2) + (y1-y2)*(y1-y2) ) ; b = sqrt( (x2-x3)*(x2-x3) + (y2-y3)*(y2-y3) ) ; c = sqrt( (x1-x3)*(x1-x3) + (y1-y3)*(y1-y3) ) ; p = ( a + b + c ) / 2.0 ; s = sqrt( p * (p-a) * (p-b) * (p-c) ) ;//三角形的面积,数学中的海伦公式 r = a * b * c / ( 4 * s ) ;//求外接圆半径 if( a > c ) { k = a ; a = c ; c = k ; } if( b > c ) { k = b ; b = c ; c = k ; } A = 2 * asin(a/(2*r)) ; B = 2 * asin(b/(2*r)) ; C = 2 * PI - A - B ; p = gcd(A,B); p = gcd(p,C) ;//此正多边形的边数为2*PI/p; //一个小三角形的面积公式为1/2*r*r*sinp; printf("%.6lf\n", (PI*r*r*sin(p))/p ) ;//n边形的面积 return 0; }
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