I Think I Need a Houseboat

in this problem, I have this solution:

#include <stdio.h>
#include <math.h>
int main()
{
  int t,p = 1;
  scanf("%d",&t);
  while(p <= t)
  {
   double x,y;
   int i=1;
   scanf("%lf%lf",&x,&y);

   double a = (x*x + y*y)*M_PI_2;//please note this line.

   while(a > i*50)
   {
    i++;
   }
   printf("Property %d: This property will begin eroding in year %d.\n",p++,i);
  }
  printf("END OF OUTPUT.");
  return 0;
}

So, here: double a = (x*x + y*y)*M_PI_2

Obviously, this line calculates the area of the circle that pass through the point we are analyzing.

mathematically, if  x*x + y*y = r*r  and  PI*r*r = A
then PI*(x*x + y*y)=A, right?

yet the result is not accepted by the judge. Instead, to be accepted, the code must look as follows:

#include <stdio.h>
#include <math.h>
int main()
{
  int t,p = 1;
  scanf("%d",&t);
  while(p <= t)
  {
   double x,y;
   int i=1;
   scanf("%lf%lf",&x,&y);

   double r = sqrt(x*x + y*y);
   double a = r*r*1.57;

   while(a > i*50)
   {
    i++;
   }
   printf("Property %d: This property will begin eroding in year %d.\n",p++,i);
  }
  printf("END OF OUTPUT.");
  return 0;
}

1.57 being PI/2 where PI = 3.14

Now, the question is, how is it possible that using 3.14 as PI value, and calculating the square root of  a number, and then squaring the result, we can come up with a more accurate result than just making a substitution in a formula and using the arbitrary precision M_PI_2 value provided by the C++ standard?

I mean, in my understanding, the lesser operations, the more accurate the result will be, wouldn磘?
I just don磘 see any reason why I should calculate the square root and then square the result to use the number, when I can directly use the number without that precision-losing calculations.

if anyone agrees with me, please let me know. And more importantly, if anyone disagrees with me, please tell me why am I wrong about this.

• Re:Problem on HouseBoat(1005) (2221) frkstyc 2006-06-27 02:16:24