Temple of Dune
| Time Limit: 1000MS |
|
Memory Limit: 65536K |
| Total Submissions: 211 |
|
Accepted: 82 |
Description
The Archaeologists of the Current Millenium (ACM) now and then discover ancient artifacts located at the vertices of regular polygons. In general it is necessary to move one sand dune to uncover each artifact. After discovering three artifacts, the archaeologists wish to compute the minimum number of dunes that must be moved to uncover all of them.
Input
The first line of input contains a positive integer n, the number of test cases. Each test case consists of three pairs of real numbers giving the x and y coordinates of three vertices from a regular polygon.
Output
For each line of input, output a single integer stating the fewest vertices that such a polygon might have. You may assume that each input case gives three distinct vertices of a regular polygon with at most 200 vertices.
Sample Input
4
10.00000 0.00000 0.00000 -10.00000 -10.00000 0.00000
22.23086 0.42320 -4.87328 11.92822 1.76914 27.57680
156.71567 -13.63236 139.03195 -22.04236 137.96925 -11.70517
129.400249 -44.695226 122.278798 -53.696996 44.828427 -83.507917
Sample Output
4
6
23
100
Source
题目大意是给出三个点的(x,y)坐标，要求输出一个边数最小的正多边形的边数，使这三个点恰好在
这个正多边形上面。其实这个三角形和这个正多边形是共外接圆，由外接圆的圆心出发，三角形的三
条边可以把圆分成三份，每份圆弧所对应的圆心角分别为arg[0],arg[1]和arg[2]，正多边形把圆弧
分成相等的n份，每份对应的圆心角为2*pi/n。其实三角形的三个角就分别占用了若干等份正多边形
所划分的圆弧，最后也就只要求arg[0],arg[1],arg[2]和2*pi的最大公约数(gcd)即可。但是这里是
个角度都是浮点数，所以还定义一个浮点数的gcd，计算浮点数的gcd可以利用math.h的函数fmod
(x,y)表示x%y。例如3.5%0.3=0.2，x%y的结果为不超过y的一个浮点数。下面写了一个fmod(x,y)自己
的实现。
double fmod(double x, double y)
{
return x-floor(x/y)*y;
}
有了fmod函数以后，就可以用它来求gcd了！
double fgcd(double a, double b)
{
double t;
if(dblcmp(a-b) == 1) //a>b
{
t=a;
a=b;
b=t;
}
if(dblcmp(a) == 0) return b;
return fgcd(fmod(b,a),a);
}
posted on 2008-06-28 15:18
飞飞 阅读(1206)
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