c 97/4/2 : intpi2 ("2" stands for "version No.2")
c NB) To avoid overflow, n <= 46340 so that n**2 < 2**31.
c
      program intpi2
      implicit integer (a-z)
      n=40000
      call intpi(n)
      end
c
      subroutine intpi(n)
      implicit integer (a-z)
      m=0
      n1=n-1
      n2=n**2
      do 20 i=0,n-1
        k=n2-(i**2+i)
      do 10 j=0,n-1
        k=k-2*j
        if(k.gt.0) m=m+1
   10 continue
   20 continue
      write(6,900) m,n**2
  900 format(' pi/4=',i12,' / ',i12)
      end