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Attached patch adds an optional format argument to time.time(), time.clock(), time.wallclock(), time.clock_gettime() and time.clock_getres() to get the timestamp as a different format. By default, the float type is still used, but it will be possible to pass "format" to get the timestamp as a decimal.Decimal object.
Some advantages of using decimal.Decimal instead of float:
 - no loss of precision during conversion from base 2 to base 10 (converting the float to a string)
 - the resolution of the clock is stored in the Decimal object
 - for big number, Decimal doesn't loose precision
Using Decimal is also motivated by the fact than Python 3.3 has now access to clock having a resolution of 1 nanosecond: clock_gettime(CLOCK_REALTIME). Well, it doesn't mean that the clock is accurate, but Python should not loose precision just because it uses floatting point instead of integers.
About the API: I chose to add a string argument to allow maybe later to support user defined formats, or at least add new builtin formats. For example, someone proposed 128 bits float in the issue #11457.
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The internal format is:
typedef struct {
 time_t seconds;
 /* floatpart can be zero */
 size_t floatpart;
 /* divisor cannot be zero */
 size_t divisor;
 /* log10 of the clock resoltion, the real resolution is 10^resolution,
 resolution is in range [-9; 0] */
 int resolution;
} _PyTime_t;
I don't know if size_t is big enough to store any "floatpart" value: time.clock() uses a LARGE_INTEGER internally. I tested my patch on Linux 32 and 64 bits, but not yet on Windows.
The internal function encoding a timestamp to Decimal caches the resolution objects (10^resolution, common clock resolutions: 10^-6, 10^-9) in a dict, and a Context object.
I computed that 22 decimal digits should be enough to compute a timestamp of 10,000 years. Extract of my patch:
"Use 12 decimal digits to store 10,000 years in seconds + 9 decimal digits for the floating part in nanoseconds + 1 decimal digit to round correctly"
>>> str(int(3600*24*365.25*10000))
'315576000000'
>>> len(str(int(3600*24*365.25*10000)))
12
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See also the issue #11457 which is linked to this topic, but not exactly the same because it concerns a low level function (os.stat()).