I messed around in desmos for like 20 mins trying to find an algebraic solution to day 6 and made this. I feel as if this is the best way to solve the problem by graphing f(x) = (x-max_time/2)^2. Finding its y-intercept, finding its vertex (max_time/2) and then finding the minimum x needed to get the max distance by solving for max_distance = f(x).
fn main() {
let mut raw_input = include_str!("input.txt")
.lines()
.map(|f| {
f.split_ascii_whitespace()
.skip(1)
.collect::<String>()
.parse::<usize>()
.unwrap()
//for part 1, first parse then collect to a Vec<&str>
}) ;
let max_time = raw_input.next().unwrap();
let max_distance = raw_input.next().unwrap();
print!("{}", determine_range(max_time, max_distance));
//let max_times = raw_input.next().unwrap();
//let max_distances = raw_input.next().unwrap();
// let races = (0..max_times.len()).map(|i| {
// determine_range(max_times[i], max_distances[i])
// });
// let total = races.reduce(|acc,x| x*acc).unwrap();
}
fn determine_range(max_time: usize, max_dist: usize) -> f64 {
let vertex = max_time as f64/2.0;
let min_y = vertex * vertex - max_dist as f64;
let min_x = vertex - min_y.sqrt()+ 0.001;
let mut res = 2.0 * ( (vertex-0.001).floor() - min_x.ceil() + 1.0);
if vertex.fract() == 0.0 {
res+=1.0;
}
res
}
ISSUE SOLVED, THANK YOU ALL