63 lines
1.5 KiB
Ruby
63 lines
1.5 KiB
Ruby
descr=%{
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A unit fraction contains 1 in the numerator. The decimal representation of the unit fractions with denominators 2 to 10 are given:
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^(1)/_(2) = 0.5
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^(1)/_(3) = 0.(3)
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^(1)/_(4) = 0.25
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^(1)/_(5) = 0.2
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^(1)/_(6) = 0.1(6)
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^(1)/_(7) = 0.(142857)
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^(1)/_(8) = 0.125
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^(1)/_(9) = 0.(1)
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^(1)/_(10) = 0.1
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Where 0.1(6) means 0.166666..., and has a 1-digit recurring cycle. It can be seen that ^(1)/_(7) has a 6-digit recurring cycle.
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Find the value of d < 1000 for which ^(1)/_(d) contains the longest recurring cycle in its decimal fraction part.
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}
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reflexion=%{
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simulate the manual division
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10 | 7
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30 |-----------
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20 | 0,142...
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...
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}
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def _rec_cycle_frac(n,m,rests,i)
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return 0 if n==0
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# print "\t_rec_cycle_frac "
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# p [n, m, rests, i]
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if n < m
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rests[n*10]=i
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return _rec_cycle_frac(n*10,m,rests,i+1)
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end
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num=(n%m) * 10
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if not rests[num].nil?
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# puts "[RES] not rest[#{num}].nil? => #{rests[num]}; #{i}-#{rests[num]}=#{i- rests[num]}"
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return i - rests[num]
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end
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rests[num]=i
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return _rec_cycle_frac( num, m, rests, i+1 )
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end
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def size_of_rec_cycle_of_frac(n,m)
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return _rec_cycle_frac(n,m,{},0)
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end
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max=0
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best=0
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(1..1000).each do |m|
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n=size_of_rec_cycle_of_frac(1,m)
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if n>max
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best=m
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max=n
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end
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puts "1/#{m}: #{n}"
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end
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puts "#{best} #{max}"
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