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  1. MAX_K = 1000000
  2. k = 0
  3. n = 0
  4.  
  5. while k < MAX_K
  6. 	n += 1
  7. 	k += (Math.log10(n) + 1).to_i
  8. end
  9.  
  10. answer = n.to_s.reverse[(k - MAX_K), 1]
  11.  
  12. puts "answer is #{answer}  (of #{n}  #{n.to_s.length - (k - MAX_K)} th)"
  13.  
Success #stdin #stdout 0.19s 4760KB
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stdin
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Standard input is empty
stdout
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answer is 1  (of 185185  1 th)
https://ideone.com/Id69U
language:
Ruby (ruby 2.5.5)
created:
12 years ago
visibility:
secret

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