Problem:
Problem B: Monday-Saturday Prime Factors

Chief Judge's log, stardate 48642.5.
We have decided to make a problem from elementary number theory.
The problem looks like finding all prime factors of a positive integer,
but it is not.

A positive integer whose remainder divided by 7 is either 1 or 6 is called
a 7N+{1,6} number.
But as it is hard to pronounce,
we shall call it a Monday-Saturday number.

For Monday-Saturday numbers a and b,
we say a is a Monday-Saturday divisor of b
if there exists a Monday-Saturday number x
such that ax = b.
It is easy to show that
for any Monday-Saturday numbers a and b,
it holds that a is
a Monday-Saturday divisor of b
if and only if
a is a divisor of b in the usual sense.

We call a Monday-Saturday number a Monday-Saturday prime
if it is greater than 1 and has no Monday-Saturday divisors
other than itself and 1.
A Monday-Saturday number which is a prime in the usual sense
is a Monday-Saturday prime
but the converse does not always hold.
For example, 27 is a Monday-Saturday prime
although it is not a prime in the usual sense.
We call a Monday-Saturday prime
which is a Monday-Saturday divisor of a Monday-Saturday number a
a Monday-Saturday prime factor of a.
For example, 27 is one of the Monday-Saturday prime factors of 216,
since 27 is a Monday-Saturday prime
and 216 = 27 × 8 holds.

Any Monday-Saturday number greater than 1
can be expressed as a product of one or more Monday-Saturday primes.
The expression is not always unique
even if differences in order are ignored.
For example,

216 = 6 × 6 × 6 = 8 × 27

holds.

Our contestants should write a program that outputs
all Monday-Saturday prime factors
of each input Monday-Saturday number.

Input

The input is a sequence of lines each of which contains a single
Monday-Saturday number.
Each Monday-Saturday number is greater than 1
and less than 300000 (three hundred thousand).
The end of the input is indicated by a line
containing a single digit 1.

Output

For each input Monday-Saturday number,
it should be printed, followed by a colon `:'
and the list of its Monday-Saturday prime factors on a single line.
Monday-Saturday prime factors should be listed in ascending order
and each should be preceded by a space.
All the Monday-Saturday prime factors should be printed only once
even if they divide the input Monday-Saturday number more than once.

Sample Input

205920
262144
262200
279936
299998
1

Output for the Sample Input

205920: 6 8 13 15 20 22 55 99
262144: 8
262200: 6 8 15 20 50 57 69 76 92 190 230 475 575 874 2185
279936: 6 8 27
299998: 299998

Solution in Rust (judge verdict: Accepted)

#[allow(unused_imports)]
use std::cmp::*;
#[allow(unused_imports)]
use std::collections::*;
use std::io::Read;
#[allow(dead_code)]
fn getline() -> String {
    let mut ret = String::new();
    std::io::stdin().read_line(&mut ret).ok().unwrap();
    ret
}
fn get_word() -> String {
    let mut stdin = std::io::stdin();
    let mut u8b: [u8; 1] = [0];
    loop {
        let mut buf: Vec<u8> = Vec::with_capacity(16);
        loop {
            let res = stdin.read(&mut u8b);
            if res.unwrap_or(0) == 0 || u8b[0] <= b' ' {
                break;
            } else {
                buf.push(u8b[0]);
            }
        }
        if buf.len() >= 1 {
            let ret = String::from_utf8(buf).unwrap();
            return ret;
        }
    }
}

#[allow(dead_code)]
fn get<T: std::str::FromStr>() -> T { get_word().parse().ok().unwrap() }

fn solve() {
    let n = 301000;
    let mut pr = vec![true; n];
    let mut fac = vec![Vec::new(); n];
    for i in 6 .. n {
        if i % 7 != 1 && i % 7 != 6 {
            pr[i] = false;
            continue;
        }
        if !pr[i] { continue; }
        fac[i].push(i);
        for j in 2 .. (n - 1) / i + 1 {
            pr[i * j] = false;
            fac[i * j].push(i);
        }
    }
    for a in fac.iter_mut() {
        a.sort();
    }
    loop {
        let u: usize = get();
        if u == 1 { break; }
        print!("{}:", u);
        for a in &fac[u] {
            print!(" {}", a);
        }
        println!();
    }
}

fn main() {
    // In order to avoid potential stack overflow, spawn a new thread.
    let stack_size = 104_857_600; // 100 MB
    let thd = std::thread::Builder::new().stack_size(stack_size);
    thd.spawn(|| solve()).unwrap().join().unwrap();
}