Arbitrary precision integer arithmetic: ceiling?

[Robert Kern]

Alasdair wrote:
> I need to apply the ceiling function to arbitrary sized (long) integers. 
> However, division automatically returns the type of its operands, so that,
> for example: math.ceil(7/4) returns 1.  I can use float, as in:
> math.ceil(7/float(4)), except that for very large integers float causes an
> unacceptable loss of precision.
> 
> What is the best way of finding a ceiling of a quotient of arbitrary sized
> integers?

Use divmod() to get the quotient and the remainder at the same time. Add 1 to 
the quotient if and only the remainder is greater than 0.


In [11]: def qceil(x, y):
    ....:     """ Find the ceiling of a quotient x/y.
    ....:
    ....:     This is especially useful for very large Python long integers.
    ....:     """
    ....:     q, r = divmod(x, y)
    ....:     if r > 0:
    ....:         q += 1
    ....:     return q
    ....:

In [13]: qceil(7, 4)
Out[13]: 2

In [14]: qceil(8, 4)
Out[14]: 2

In [15]: qceil(9, 4)
Out[15]: 3

In [16]: qceil(100000000000000000000000003, 10)
Out[16]: 10000000000000000000000001L

-- 
Robert Kern

"I have come to believe that the whole world is an enigma, a harmless enigma
  that is made terrible by our own mad attempt to interpret it as though it had
  an underlying truth."
   -- Umberto Eco