結果

問題 No.1644 Eight Digits
ユーザー terry_u16
提出日時 2021-08-13 22:37:10
言語 Rust
(1.83.0 + proconio)
結果
AC  
実行時間 7 ms / 1,000 ms
コード長 4,781 bytes
コンパイル時間 13,453 ms
コンパイル使用メモリ 386,744 KB
実行使用メモリ 5,248 KB
最終ジャッジ日時 2024-12-21 22:51:12
合計ジャッジ時間 15,050 ms
ジャッジサーバーID
（参考情報） 		judge4 / judge2
このコードへのチャレンジ
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ファイルパターン 結果
sample AC * 3
other AC * 25
権限があれば一括ダウンロードができます

ソースコード

diff #
プレゼンテーションモードにする

pub trait ChangeBest {
    fn chmin(&mut self, v: Self) -> bool;
    fn chmax(&mut self, v: Self) -> bool;
}
impl<T> ChangeBest for T
where
    T: PartialOrd,
{
    fn chmin(&mut self, v: T) -> bool {
        if *self > v {
            *self = v;
            true
        } else {
            false
        }
    }
    fn chmax(&mut self, v: T) -> bool {
        if *self < v {
            *self = v;
            true
        } else {
            false
        }
    }
}
#[allow(unused_macros)]
macro_rules! mat {
    ($e:expr; $d:expr) => { vec![$e; $d] };
    ($e:expr; $d:expr $(; $ds:expr)+) => { vec![mat![$e $(; $ds)*]; $d] };
}
#[allow(unused_macros)]
macro_rules! input {
    (source = $s:expr, $($r:tt)*) => {
        let mut iter = $s.split_whitespace();
        let mut next = || { iter.next().unwrap() };
        input_inner!{next, $($r)*}
    };
    ($($r:tt)*) => {
        let stdin = std::io::stdin();
        let mut bytes = std::io::Read::bytes(std::io::BufReader::new(stdin.lock()));
        let mut next = move || -> String{
            bytes
                .by_ref()
                .map(|r|r.unwrap() as char)
                .skip_while(|c|c.is_whitespace())
                .take_while(|c|!c.is_whitespace())
                .collect()
        };
        input_inner!{next, $($r)*}
    };
}
#[allow(unused_macros)]
macro_rules! input_inner {
    ($next:expr) => {};
    ($next:expr, ) => {};
    ($next:expr, $var:ident : $t:tt $($r:tt)*) => {
        let $var = read_value!($next, $t);
        input_inner!{$next $($r)*}
    };
}
#[allow(unused_macros)]
macro_rules! read_value {
    ($next:expr, ( $($t:tt),* )) => {
        ( $(read_value!($next, $t)),* )
    };
    ($next:expr, [ $t:tt ; $len:expr ]) => {
        (0..$len).map(|_| read_value!($next, $t)).collect::<Vec<_>>()
    };
    ($next:expr, chars) => {
        read_value!($next, String).chars().collect::<Vec<char>>()
    };
    ($next:expr, usize1) => {
        read_value!($next, usize) - 1
    };
    ($next:expr, $t:ty) => {
        $next().parse::<$t>().expect("Parse error")
    };
}
pub trait LexicalPermutation {
    /// Return `true` if the slice was permuted, `false` if it is already
    /// at the last ordered permutation.
    fn next_permutation(&mut self) -> bool;
    /// Return `true` if the slice was permuted, `false` if it is already
    /// at the first ordered permutation.
    fn prev_permutation(&mut self) -> bool;
}
impl<T> LexicalPermutation for [T] where T: Ord {
    /// Original author in Rust: Thomas Backman <serenity@exscape.org>
    fn next_permutation(&mut self) -> bool {
        // These cases only have 1 permutation each, so we can't do anything.
        if self.len() < 2 { return false; }
        // Step 1: Identify the longest, rightmost weakly decreasing part of the vector
        let mut i = self.len() - 1;
        while i > 0 && self[i-1] >= self[i] {
            i -= 1;
        }
        // If that is the entire vector, this is the last-ordered permutation.
        if i == 0 {
            return false;
        }
        // Step 2: Find the rightmost element larger than the pivot (i-1)
        let mut j = self.len() - 1;
        while j >= i && self[j] <= self[i-1]  {
            j -= 1;
        }
        // Step 3: Swap that element with the pivot
        self.swap(j, i-1);
        // Step 4: Reverse the (previously) weakly decreasing part
        self[i..].reverse();
        true
    }
    fn prev_permutation(&mut self) -> bool {
        // These cases only have 1 permutation each, so we can't do anything.
        if self.len() < 2 { return false; }
        // Step 1: Identify the longest, rightmost weakly increasing part of the vector
        let mut i = self.len() - 1;
        while i > 0 && self[i-1] <= self[i] {
            i -= 1;
        }
        // If that is the entire vector, this is the first-ordered permutation.
        if i == 0 {
            return false;
        }
        // Step 2: Reverse the weakly increasing part
        self[i..].reverse();
        // Step 3: Find the rightmost element equal to or bigger than the pivot (i-1)
        let mut j = self.len() - 1;
        while j >= i && self[j-1] < self[i-1]  {
            j -= 1;
        }
        // Step 4: Swap that element with the pivot
        self.swap(i-1, j);
        true
    }
}
fn main() {
    input! {
        k: i32
    };
    let mut s = "12345678".chars().collect::<Vec<_>>();
    let mut set = std::collections::HashSet::new();
    s.sort_unstable();
    loop {
        let n = s.iter().collect::<String>().parse::<i32>().unwrap();
        if n % k == 0 {
            set.insert(n);
        }
        if !s.next_permutation() {
            break;
        }
    }
    println!("{}", set.len());
}
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