Exercise 1 solution

    from datetime import datetime, timezone, timedelta

    EPOCH = datetime(1970, 1, 1, tzinfo=timezone.utc)

    def to_seconds(v):
        a = abs(v)
        if a < 1e11:
            return v
        if a < 1e14:
            return v / 1e3
        if a < 1e17:
            return v / 1e6
        return v / 1e9

    def to_dt(v):
        return EPOCH + timedelta(seconds=to_seconds(v))

    stamps = [1_793_620_800, 1_793_620_800_500, 1_793_624_400_000,
              1_793_706_000_250_000, 0, -86_400]
    for v in stamps:
        print(v, "->", to_dt(v).isoformat())
    print("Elapsed:", to_dt(stamps[2]) - to_dt(stamps[1]))
    print("Elapsed:", to_dt(stamps[3]) - to_dt(stamps[0]))

Output (run, and exactly as shown):

    1793620800 -> 2026-11-02T12:00:00+00:00
    1793620800500 -> 2026-11-02T12:00:00.500000+00:00
    1793624400000 -> 2026-11-02T13:00:00+00:00
    1793706000250000 -> 2026-11-03T11:40:00.250000+00:00
    0 -> 1970-01-01T00:00:00+00:00
    -86400 -> 1969-12-31T00:00:00+00:00
    Elapsed: 0:59:59.500000
    Elapsed: 23:40:00.250000

Explanation: values under 1e11 are seconds, under 1e14 milliseconds, under
1e17 microseconds, otherwise nanoseconds. These bounds cover dates from 1970
to well past the year 5000 in seconds, and the digit count of a current
timestamp tells you its unit: 10 digits seconds, 13 milliseconds, 16
microseconds, 19 nanoseconds. Zero and negative values are seconds, and
adding a timedelta to the epoch handles negatives on every platform, whereas
fromtimestamp can fail on some systems for dates before 1970. Subtracting two
aware datetimes gives the elapsed time directly.