These might be helpful:
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thanks, but i can’t find the solution, in a codeforces problem i need to calculate ceil of a big number, how can i do that?
By not using float.
You should use decimal.Decimal
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i wonder if there is a way to set in python to use this by default?
I don’t think there is or there should be. float is faster and lighter.
If you really need it to be precise, you have options.
This is a common problem on almost every other language.
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And then they might have the same problem just a little later.
Really they should tells us more about what they’re doing, so that we can suggest a real proper solution.
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I don’t see anything “incorrect” here. On your first example you have something like this:
>>> 1.00000000000000001
1.0
>>> math.ceil(_) # (1)
1
>>> 1.0000000001
1.0000000001
>>> math.ceil(_) # (2)
2
Probably, you expected (2) in both cases. But floating-point numbers have a fixed precision. So, if you put enough zeros before trailing 1 — the value will be eventually rounded to 1.0. It’s float with an integer value (floor() equals to cell()).
Similarly your last example:
>>> 2832139218378291759861231213.1 # (3)
2.8321392183782917e+27
>>> _.is_integer()
True
>>> math.ceil(2832139218378291759861231213.1)
2832139218378291710804361216
You put too much digits in, so your input was rounded and you get some unexpected (for you) floating-point integer value at (3).
Indeed.
You will have similar with the decimal module, though it’s default precision is bigger (28 decimal significant digits vs 15-17 for floats):
>>> import decimal, math
>>> decimal.getcontext().prec = 5
>>> +decimal.Decimal('1.00000000001')
Decimal('1.0000')
>>> math.ceil(+decimal.Decimal('1.00000000001'))
1
>>> +decimal.Decimal('123445.1')
Decimal('1.2345E+5')
>>> math.ceil(+decimal.Decimal('123445.1'))
123450
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