WM. wrote:
# The natural numbers(natnum), under 1000, divisible by 3 or by 5 are
to be added together.
natnum = 0
num3 = 0
num5 = 0
cume = 0
# The 'and' is the 15 filter; the 'or' is the 3 or 5 filter.
while natnum <= 999:
num3 = natnum/3
num5 = natnum/5
if natnum - (num3 * 3) == 0 and natnum - (num5 * 5) == 0:
cume = cume + natnum
elif natnum - (num3 * 3) == 0 or natnum - (num5 * 5) == 0:
if natnum - (num3 * 3) == 0:
cume = cume + natnum
elif natnum - (num5 * 5) == 0:
cume = cume + natnum
natnum = natnum + 1
print cume
My head hurts trying to follow all that logic and expressions. That is a
sign of too much code. Let's simplify:
1) no need to initialize num3 or num5
2) replace while with for
3a) replace natnum - (num3 * 3) == 0 with natnum == (num3 * 3)
3b) replact that with not (natnum % 3)
4) use += and -=
for natnum in range(1000):
if not (natnum % 3 or natnum % 5): # adds 1 if divisible by 3 and/or 5
cume += 1
if not natnum % 15: # remove one of the ones if divisible by 15
cume -= 1
Since the result of not is True (1) or False (0) one can also write
cume += not (natnum % 3 or natnum % 5)
cume 1= not (natnum % 15)
An alternative:
for natnum in range(1000):
if not natnum % 15:
pass
elif not (natnum % 3 or natnum % 5): # adds 1 if divisible by 3 or 5
cume += 1
This problem was kicked around last month and I did not understand any
of the scripts.
That's really unfortunate. Would you for your learning take one of them,
tell us what if any parts of it you understand and let us nudge you
toward more understanding.
--
Bob Gailer
Chapel Hill NC
919-636-4239
_______________________________________________
Tutor maillist - Tutor@python.org
http://mail.python.org/mailman/listinfo/tutor
