In article <4bf442cd$0$31377$4fafb...@reader1.news.tin.it>,
superpollo <ute...@esempio.net> wrote:
>... how many positive integers less than n have digits that sum up to m:
>
>In [197]: def prttn(m, n):
> tot = 0
> for i in range(n):
> s = str(i)
> sum = 0
> for j in range(len(s)):
> sum += int(s[j])
> if sum == m:
> tot += 1
> return tot
> .....:
>
>In [207]: prttn(25, 10000)
>Out[207]: 348
>
>any suggestion for pythonizin' it?
I don't like the excursion to string and back.
def x(i) : return x(i/10)+i%10 if i else 0
or if you can't stand recursion:
def x(i):
s= 0
while i:
s += i%10
i /= 10
return s
(All untested but you get the idea.)
>
>bye
--
--
Albert van der Horst, UTRECHT,THE NETHERLANDS
Economic growth -- being exponential -- ultimately falters.
alb...@spe&ar&c.xs4all.nl &=n http://home.hccnet.nl/a.w.m.van.der.horst
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