Minimum positive multiple in base 10 using only 0 and 1: Difference between revisions
Minimum positive multiple in base 10 using only 0 and 1 (view source)
Revision as of 21:01, 2 November 2021
, 2 years agoPython example
m (→{{header|AppleScript}}: Understanding achieved, commenting and rationalisation completed. For now, anyway.) |
(Python example) |
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"0.1s"
</pre>
=={{header|Python}}==
{{trans|Kotlin}}
<lang python>def getA004290(n):
if n < 2:
return 1
arr = [[0 for _ in range(n)] for _ in range(n)]
arr[0][0] = 1
arr[0][1] = 1
m = 0
while True:
m += 1
if arr[m - 1][-10 ** m % n] == 1:
break
arr[m][0] = 1
for k in range(1, n):
arr[m][k] = max([arr[m - 1][k], arr[m - 1][k - 10 ** m % n]])
r = 10 ** m
k = -r % n
for j in range((m - 1), 0, -1):
if arr[j - 1][k] == 0:
r = r + 10 ** j
k = (k - 10 ** j) % n
if k == 1:
r += 1
return r
for n in [i for i in range(1, 11)] + \
[i for i in range(95, 106)] + \
[297, 576, 594, 891, 909, 999, 1998, 2079, 2251, 2277, 2439, 2997, 4878]:
result = getA004290(n)
print(f"A004290({n}) = {result} = {n} * {result // n})")
</lang>{{out}} Same as Kotlin.
=={{header|Raku}}==
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