33 lines
834 B
Python
33 lines
834 B
Python
from lib import *
|
|
|
|
row = open(0).read().strip()
|
|
|
|
def is_trap(s):
|
|
assert len(s) == 3
|
|
match s:
|
|
case "^^.":
|
|
return "^"
|
|
case ".^^":
|
|
return "^"
|
|
case "^..":
|
|
return "^"
|
|
case "..^":
|
|
return "^"
|
|
case _:
|
|
return "."
|
|
|
|
# 400_000 is still easily bruteforcible. If it was a much larget number, we
|
|
# would have to cache the rows till we get a repeated row. We would then
|
|
# memorize the number of safe tiles from repeated row to repeated row and jump
|
|
# forward by multiples of that amount to reach much higher numbers.
|
|
r = 0
|
|
for _ in range(400_000):
|
|
r += row.count(".")
|
|
nrow = is_trap("." + row[:2])
|
|
for i in range(1, len(row) - 1):
|
|
nrow += is_trap(row[i-1:i+2])
|
|
nrow += is_trap(row[-2:] + ".")
|
|
row = nrow
|
|
|
|
print(r)
|