Function aoc::y22d10::y22d10p1

pub fn y22d10p1(input: &str) -> i32

The solution for part one of the day ten challenge.

This is relatively straightforward after computing the state of the register for each cycle. It just involves a little bit of math. The signal strength is computed by the value of the register at cycle 20 and then again every 40 cycles thereafter. So we start by calculating how many additional checks we need to make (the length of the cycles vector minus 20 to account for the first check divided by 40). Then for each check we add to the signal strength the value of the register times the cycle (index/number).

Example

// probably read this from the input file...
let input = "noop\naddx 3\naddx -5";
assert_eq!(y22d10p1(input), 0);