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  1. n=9
  2. d=1
  3. pos=int(input())
  4. while True:
  5. 	sub=n*d
  6. 	if sub <= pos:
  7. 		pos-=sub
  8. 	else:
  9. 		break
  10. 	n*=10
  11. 	d+=1
  12. res=10**(d-1)
  13. res+=(pos//d)-(1 if pos%d==0 else 0)
  14. print((d if pos%d==0 else pos%d), end='')
  15. print("th digit of ",res," which is ", end='')
  16. pos_f=d-(d if (pos%d==0) else pos%d)
  17. while pos_f>0:
  18. 	res//=10
  19. 	pos_f-=1
  20. print(res%10)
Success #stdin #stdout 0.02s 9936KB
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stdin
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999999999
stdout
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9th digit of  123456789  which is 9
https://ideone.com/fUikKD
Python equivalent of code uOpELp
language:
Python 3 (python  3.12)
created:
9 years ago
visibility:
public

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