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  1. def f(n):
  2.   return int((20130000*n-2013.5)/20129999)
  3.  
  4. n=1
  5. p=f(n)
  6. s=f(n+1)
  7.  
  8. while s==p+1:
  9.   p=s
  10.   n+=1
  11.   s=f(n+1)
  12.  
  13. print(n)
Success #stdin #stdout 0.02s 5760KB
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2013
https://ideone.com/Wvij6Q
language:
Python 3 (python  3.9.5)
created:
11 years ago
visibility:
secret

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