The editor gets the numbers on the entrance and transforms it. The editor can perform two teams, both teams v and w mark the symbol chains
If I understood algorithm correctly, the code looks like:
i = 300 while True: i += 1
text = '5' * i while '555' in text or '888' in text: index = text.find('555') text = text if index == -1 else (text[:index] + '8' + text[index + 3:]) index = text.find('888') text = text if index == -1 else (text[:index] + '55' + text[index + 3:]) count5 = text.count('5') count8 = text.count('8') if count5 > count8: print(i) break
and response 303
It is theoretically possible to decide again:
It is clear that, with a change only on the left to the right, we can only get the following options:
(no versions of type 858... because we're changing numbers on the left to the right and there's no way
'5'- There's nowhere to build it.
'888'Always changing strictly.
Of which, by condition, we're only welcome.
55(G8 and 5 from 0 to 2 are defined precisely by the fact that we replace three of these numbers, i.e. model 3).
How hard to understand all the lines above will be repeated with the period 7
Now that we have a periodicity to start from scratch, we'll look at the lines of one symbol and above:
'5'We don't like it, it's out of seven options.
We' the next line.
'55'- It's size 2 now.
2 + 7 * x > 300
x = 47
2 + 7 * 47 = 303
Target accomplished without programming