19 lines
325 B
Text
19 lines
325 B
Text
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F reps(text)
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R (1 .< 1 + text.len I/ 2).filter(x -> @text.starts_with(@text[x..])).map(x -> @text[0 .< x])
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V matchstr =
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|‘1001110011
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1110111011
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0010010010
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1010101010
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1111111111
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0100101101
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0100100
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101
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11
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00
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1’
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L(line) matchstr.split("\n")
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print(‘'#.' has reps #.’.format(line, reps(line)))
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