Codeforces beta round #1 exercise

Question

B. Exercise caution when handling the simulation questions.

 

 

C. Give you three points on a positive polygon and determine the minimum area of the polygon.

Since this polygon has a maximum of 100 edges, you can calculate whether the number of edges of the polygon can be satisfied.

Determine whether the conditions are met:

Each angle of a triangle composed of three points is an integer multiple of the circumference angle.

The precision should not be too big when determining the integer

View code

# Include <cstdio> # Include <Cstring> # Include <Cmath> # Include <Cstdlib> Using   Namespace  STD;  Const  Double EPS = 1E- 8  ;  Const   Double Pi = ACOs (- 1.0  );  Double Dist ( Double X1, Double Y1, Double X2, Double  Y2 ){  Return SQRT (x1-x2) * (x1-x2) + (y1-y2) * (Y1-Y2 ));}  Double Cross ( Double X1, Double Y1, Double X2, Double Y2, Double X3, Double  Y3 ){  Return (X2-x1) * (y3-y1)-(x3-x1) * (Y2- Y1 );}  Bool OK ( Int N, Double Ang ){  Double TMP = ang * n/PI; //  Determines whether TMP is an integer.      Double X = floor (TMP + 1E- 3 ); //  The precision should not be too large. Here we wa it once      If (TMP-x <1E- 3 ) Return   True  ;  Return   False ;}  Int  Main (){  Double  X1, x2, Y1, Y2, X3, Y3;  While (Scanf ( "  % Lf  " , & X1, & Y1, & X2, & Y2, & X3, & Y3 )! = EOF ){  Double A = Dist (x1, Y1, X2, Y2 );  Double B = Dist (X2, Y2, X3, Y3 ); Double C = Dist (x1, Y1, X3, Y3 );  Double S = Cross (x1, Y1, X2, Y2, X3, Y3 );  Double R = a * B * C/ 2 / S;  Double C = ACOs (A * A + B * B-c * C )/( 2 * * B ));  Double B = ACOs (A * A + C * C-B * B )/( 2 * * C )); Double A = ACOs (B * B + C * C-A * )/( 2 * B * C ));  //  Printf ("%. 2lf %. 2lf %. 2lf \ n", A, B, C );          Int  N;  For (N = 3 ; N <= 100 ; N ++ )  If (OK (n, a) & OK (n, B )&& OK (N, C ))  Break ;  //  Printf ("n = % d \ n", N );          Double CEN = 2 * PI/ N;  Double Ans = r * 0.5 * Sin (CEN )* N; printf (  "  % Lf \ n  "  , ANS );}  Return   0 ;}