10432 - Polygon Inside A Circle

[Joseph Kurniawan]

Thanx for tip, boss!
Muchas gracias!!

[Joseph Kurniawan]

I have an another question.
Suppose the number of sides is 2000, then the circle will be divided in 2000 parts which will make the angle of every part 360/2000 degree.
So to calculate the sin is:

Code: Select all

pi=3.141593;
angle=sin(2*pi/n);

Am I correct?

[Faizur]

I think you should better use pi=2*acos(0)

[Joseph Kurniawan]

Thanx dude!!!
Finally got it AC in this darn prob!!

[Joseph Kurniawan]

Now I got another conversion prob.
Can anyone tell me how to convert minutes to degrees?
Thanx in advance.

[UFP2161]

1 (arc) second = 1/60 (arc) minute = 1/60 degree

[Observer]

should it be:
1 deg = 60 min = 3600 sec ?

[Faizur]

I think so...

[Joseph Kurniawan]

I've tried both

1 (arc) second = 1/60 (arc) minute = 1/60 degree

and

1 deg = 60 min = 3600 sec

and find the second is the correct one.
Thanx anyway for the help guys!!!

[WR]

Hi,

I'm getting a WA for this problem.

First I tried a straightforward approach:
- divide the polygone into triangles,
- compute height and base of a triangle
- compute the area of a triangle
- multiply by no. of triangles to get the area of the polygone.

---> WA

After that I tried the equations I've found here.

---> WA

Subsequently I created a file containing 5000 random values in the
range 1 .. 20000 and ran all four versions.

Identical results but WA.

Code: Select all

int main(void)
{
  int r, n;
  double area, dn, dr;
  double al, h, x, x2, b;
  double PI;
  while (scanf("%lf %d",&dr,&n) == 2){
    dn = (double) n;
/*    dr = (double) r;*/
    printf("\n");

/* 1st version: WA
   compute height and base for a triangle,
   find area and multiply by n
    PI = 3.1415926535897932384626;
    al = 180.0/dn;
    x = 2.0*dr*sin(al*PI/180.0);
    x2 = x*0.5;
    h = sqrt(dr*dr-x2*x2);
    area = dn*(x*h*0.5);
    printf("%.3lf\n",area);*/

/* 2nd version: WA
   simplified equation from a post
    PI = 3.1415926535897932384626433832795029L;
    al = 180.0 - ((double)180.0*(dn-2.0)/dn);
    area = (sin(PI/180.0*al)*dr*dr*dn)/2.0;
    printf("%.3lf\n",area);  */

/* 3rd version: WA
   same as eq. 2 with some constants replaced 
    PI = 2.0*acos(0.0);
    b = acos(-1.0);
    al = b - (b*(dn-2.0)/dn);
    area = (sin(PI/b*al)*dr*dr*dn)*0.5;
    printf("%.3lf\n",area);    */

/* 4th version: WA
   same as 3 with new eq. for sin(alpha) */
    al = sin(acos(-1.0)*2.0/n);
    area = (al*dr*dr*dn)*0.5;
    printf("%.3lf\n",area);
  }
  return 0;
}

Hints and/or test data would be very welcome!

Thanks

[Dominik Michniewski]

I use 4th way and got Acc ...

Best regards
DM

[WR]

Yes, me too, thanks a lot.

But I'm still asking myself why the first three solutions
resulted in a WA.

As I said, they produced absolutely identical results
(maybe it's a Windows -- Linux problem?). No idea!
But leaves one wondering how to solve other geometrical
problems.

Ok, thanks again.

[Dani Rodrigo]

I received WA many times. I don't know the reason, but finally I read numbers as doubles and got AC.

[p!ter]

Hi,
Can someont tell me what is wrong with my code.
Outputs seems ok

Code: Select all

Code removed after accepted

Thanks in advance!!!!!!

[sohel]

Radius may not always be integers.. it might take real values.

and it is safer to use acos(-1) for PI rather than 3.1415.......