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Sum multiples of 3 and 5: Difference between revisions
→{{header|Perl 6}}: Add Python. (three ways).
Walterpachl (talk | contribs) (→{{header|REXX}}: changed to below :-) thanks) |
(→{{header|Perl 6}}: Add Python. (three ways).) |
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<pre>233168</pre>
=={{header|Python}}==
Three ways of performing the calculation are shown including direct calculation of the value without having to do explicit sums in sum35c()
<lang python>def sum35a(n):
'Direct count'
# note: ranges go to n-1
return sum(x for x in range(n) if x%3==0 or x%5==0)
def sum35b(n):
"Count all the 3's; all the 5's; minus double-counted 3*5's"
# note: ranges go to n-1
return sum(range(3, n, 3)) + sum(range(5, n, 5)) - sum(range(15, n, 15))
def sum35c(n):
'Sum the arithmetic progressions: sum3 + sum5 - sum15'
consts = (3, 5, 15)
# Note: stop at n-1
divs = [(n-1) // c for c in consts]
sums = [d*c*(1+d)/2 for d,c in zip(divs, consts)]
return sum(sums[:-1]) - sums[-1]
#test
for n in range(1001):
sa, sb, sc = sum35a(n), sum35b(n), sum35c(n)
assert sa == sb and sa == sc
print('For n = %i -> %i' % (n, sc))</lang>
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<pre>For n = 1000 -> 233168</pre>
=={{header|REXX}}==
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