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
題目大意是給出三個點的(x,y)坐標，要求輸出一個邊數最小的正多邊形的邊數，使這三個點恰好在
這個正多邊形上面。其實這個三角形和這個正多邊形是共外接圓，由外接圓的圓心出發，三角形的三
條邊可以把圓分成三份，每份圓弧所對應的圓心角分別為arg[0],arg[1]和arg[2]，正多邊形把圓弧
分成相等的n份，每份對應的圓心角為2*pi/n。其實三角形的三個角就分別占用了若干等份正多邊形
所劃分的圓弧，最后也就只要求arg[0],arg[1],arg[2]和2*pi的最大公約數(gcd)即可。但是這里是
個角度都是浮點數，所以還定義一個浮點數的gcd，計算浮點數的gcd可以利用math.h的函數fmod
(x,y)表示x%y。例如3.5%0.3=0.2，x%y的結果為不超過y的一個浮點數。下面寫了一個fmod(x,y)自己
的實現。
double fmod(double x, double y)
{
return x-floor(x/y)*y;
}
有了fmod函數以后，就可以用它來求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
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