from __future__ import division
for x in xrange(30):
  J = 7205759403792790 + x
  print J * 10**30 // 2**56

T = 0.1 # Target
# T ~= J / (2**N)
# where J has exactly R bits
# J ~= T * (2**N)
print "BEst",T*(2**52)
print "    ",T*(2**53)

import math


print 52 - math.log(T,2)
print 53 - math.log(T,2)

def precision_approx_equal(self,other,precision=24):
  # Use concepts from IEEE-754 to determine if the difference between
  # two numbers is within floating point error.  Not within scope to
  # do this for double precision literals; only interested in the case
  # where the data are from a ~single precision digital-analog converter.
  if self==other:
    return True
  if (self > 0.) != (other > 0.) :
    return False
  #compute the exponent
  import math
  T = abs(self)
  Np = math.floor(precision-math.log(T,2))
  significand = int(T * 2**Np)
  val1 = significand/(2**Np) # nearest floating point representation of self
  val2 = (significand+1)/(2**Np) # next-nearest
  return abs(T-abs(other)) <= abs(val1-val2)
print precision_approx_equal(0.799999,0.800004,precision=17)