Function advent_solutions::advent2017::day17::part1

pub fn part1(input: &usize) -> usize

For example, if the spinlock were to step 3 times per insert, the circular buffer would begin to evolve like this (using parentheses to mark the current position after each iteration of the algorithm):

• (0), the initial state before any insertions.
• 0 (1): the spinlock steps forward three times (0, 0, 0), and then inserts the first value, 1, after it. 1 becomes the current position.
• 0 (2) 1: the spinlock steps forward three times (0, 1, 0), and then inserts the second value, 2, after it. 2 becomes the current position.
• 0 2 (3) 1: the spinlock steps forward three times (1, 0, 2), and then inserts the third value, 3, after it. 3 becomes the current position.

And so on:

• 0 2 (4) 3 1
• 0 (5) 2 4 3 1
• 0 5 2 4 3 (6) 1
• 0 5 (7) 2 4 3 6 1
• 0 5 7 2 4 3 (8) 6 1
• 0 (9) 5 7 2 4 3 8 6 1

Eventually, after 2017 insertions, the section of the circular buffer near the last insertion looks like this:

1512  1134  151 (2017) 638  1513  851

Perhaps, if you can identify the value that will ultimately be after the last value written (2017), you can short-circuit the spinlock. In this example, that would be 638.

assert_eq!(part1(&3), 638)

What is the value after 2017 in your completed circular buffer?