Cleanup blank lines

Author: maneatingape

src/year2015/day18.rs

@@ -65,7 +65,6 @@
 //!
 //! Using the bits as labelled above, the next cell is `(e|s) & (q^r) & !p`, masked back
 //! to the 50 bits per integer.
-
 const MASK: u64 = (1 << 50) - 1;
 const LEFT_CORNER: u64 = 1 << 49;
 const RIGHT_CORNER: u64 = 1 << 0;

src/year2016/day13.rs

@@ -8,7 +8,6 @@
 //! part one requires more than 50 steps but should easily be reachable without exceeding bounds,
 //! we can bound the maze to 2 + 50 = 52 in each dimension and use a fixed-size array. Rather
 //! than filling the array up front, we can lazily populate it as the horizon expands.
-
 use crate::util::parse::*;
 use std::collections::VecDeque;
 

src/year2020/day06.rs

@@ -14,7 +14,6 @@
 //!
 //! [`day 5`]: crate::year2020::day05
 //! [`count_ones`]: u32::count_ones
-
 pub fn parse(input: &str) -> Vec<u32> {
     input.lines().map(|line| line.bytes().fold(0, |acc, b| acc | (1 << (b - b'a')))).collect()
 }

src/year2022/day08.rs

@@ -2,7 +2,6 @@
 //!
 //! Part one is solved with an efficient `O(n)` algorithm. Part two is also solved with an efficient `O(n)`
 //! algorithm, using a bit manipulation trick to make the complexity independent of the number of digits.
-
 const ONES: u64 = 0x0041041041041041;
 const MASK: u64 = 0x0fffffffffffffc0;