phi = (1 + 5**0.5) / 2

 def fib(n):

    return int(round((phi**n - (1-phi)**n) / 5**0.5))

 for i in range(0,50):
    if 1234567890<= fib(i) and fib(i) <= 9876543210:
       print(i)

Not that elegant but very fast code.....

def fib(n):

  f1=1
  f2=1
  if n==1 or n==2:
      return 1
 else:
     for i in range(3,n+1):
         f1,f2=f2,f1+f2
 return f2



 count=0
 for i in range(1,200):
       if fib(i)>=1234567890 and fib(i)<=9876543210:
      count+=1
 print count

4

My code:

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a = 1
b = 1
N = 0

while b < 9876543211:
  if b > 1234567889:
    N += 1
  temp = a+b
  a = b
  b = temp

print(N)

python 3.4:

Solution using Haskell:

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> let fibs = 1 : 1 : zipWith (+) fibs (tail fibs)
> takeWhile (<=9876543210) $ dropWhile (<1234567890) fibs
> [1836311903,2971215073,4807526976,7778742049]

So the answer is $4$

Here is how I coded it up

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def fib(n):
    k = 0
    m = 1
    for i in range(n):
        sub = m
        m = k+m
        k = sub
    return k

def fib_in_range(a,b):
    n=0
    num = 0
    while fib(n) <= b:
        if fib(n) >= a:
            num +=1
        n += 1
    return num

print(fib_in_range(1234567890,9876543210))