結果

問題 No.694 square1001 and Permutation 3
ユーザー koba-e964
提出日時 2022-01-04 20:28:57
言語 Rust
(1.83.0 + proconio)
結果
AC  
実行時間 645 ms / 3,000 ms
コード長 4,285 bytes
コンパイル時間 11,818 ms
コンパイル使用メモリ 380,152 KB
実行使用メモリ 33,240 KB
最終ジャッジ日時 2024-10-14 16:25:47
合計ジャッジ時間 17,095 ms
ジャッジサーバーID
(参考情報)
judge4 / judge3
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 * 1
other AC * 13
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ソースコード

diff #
プレゼンテーションモードにする

use std::io::{Write, BufWriter};
// https://qiita.com/tanakh/items/0ba42c7ca36cd29d0ac8
macro_rules! input {
($($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)*}
};
}
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)*}
};
}
macro_rules! read_value {
($next:expr, [ $t:tt ; $len:expr ]) => {
(0..$len).map(|_| read_value!($next, $t)).collect::<Vec<_>>()
};
($next:expr, $t:ty) => ($next().parse::<$t>().expect("Parse error"));
}
/**
* Segment Tree. This data structure is useful for fast folding on intervals of an array
* whose elements are elements of monoid I. Note that constructing this tree requires the identity
* element of I and the operation of I.
* Verified by: yukicoder No. 259 (http://yukicoder.me/submissions/100581)
* AGC015-E (http://agc015.contest.atcoder.jp/submissions/1461001)
* yukicoder No. 833 (https://yukicoder.me/submissions/703521)
*/
struct SegTree<I, BiOp> {
n: usize,
orign: usize,
dat: Vec<I>,
op: BiOp,
e: I,
}
impl<I, BiOp> SegTree<I, BiOp>
where BiOp: Fn(I, I) -> I,
I: Copy {
pub fn new(n_: usize, op: BiOp, e: I) -> Self {
let mut n = 1;
while n < n_ { n *= 2; } // n is a power of 2
SegTree {n: n, orign: n_, dat: vec![e; 2 * n - 1], op: op, e: e}
}
/* ary[k] <- v */
pub fn update(&mut self, idx: usize, v: I) {
debug_assert!(idx < self.orign);
let mut k = idx + self.n - 1;
self.dat[k] = v;
while k > 0 {
k = (k - 1) / 2;
self.dat[k] = (self.op)(self.dat[2 * k + 1], self.dat[2 * k + 2]);
}
}
/* [a, b) (note: half-inclusive)
* http://proc-cpuinfo.fixstars.com/2017/07/optimize-segment-tree/ */
#[allow(unused)]
pub fn query(&self, mut a: usize, mut b: usize) -> I {
debug_assert!(a <= b);
debug_assert!(b <= self.orign);
let mut left = self.e;
let mut right = self.e;
a += self.n - 1;
b += self.n - 1;
while a < b {
if (a & 1) == 0 {
left = (self.op)(left, self.dat[a]);
}
if (b & 1) == 0 {
right = (self.op)(self.dat[b - 1], right);
}
a = a / 2;
b = (b - 1) / 2;
}
(self.op)(left, right)
}
}
trait Change { fn chmax(&mut self, x: Self); fn chmin(&mut self, x: Self); }
impl<T: PartialOrd> Change for T {
fn chmax(&mut self, x: T) { if *self < x { *self = x; } }
fn chmin(&mut self, x: T) { if *self > x { *self = x; } }
}
// (#inversion, #equal pairs)
fn find_acc(a: &[i64]) -> (Vec<i64>, Vec<i64>) {
let n = a.len();
let mut coo = a.to_vec();
coo.sort(); coo.dedup();
let m = coo.len();
let mut st = SegTree::new(m, |x, y| x + y, 0);
let mut inv = vec![0; n + 1];
let mut eq = vec![0; n + 1];
for i in 0..n {
let idx = coo.binary_search(&a[i]).unwrap();
inv[i + 1] = inv[i] + st.query(idx + 1, m);
let val = st.query(idx, idx + 1);
eq[i + 1] = eq[i] + val;
st.update(idx, val + 1);
}
(inv, eq)
}
fn main() {
let out = std::io::stdout();
let mut out = BufWriter::new(out.lock());
macro_rules! puts {($($format:tt)*) => (let _ = write!(out,$($format)*););}
input! {
n: usize,
a: [i64; n],
}
let (inv1, eq1) = find_acc(&a);
let mut b = a;
b.reverse();
for v in &mut b {
*v *= -1;
}
let (mut inv2, mut eq2) = find_acc(&b);
inv2.reverse();
eq2.reverse();
for i in 0..n {
let whole = i as i64 * (n - i) as i64 - (eq1[n] - eq1[i] - eq2[i]);
let rem = inv1[n] - inv1[i] - inv2[i];
puts!("{}\n", inv1[i] + inv2[i] + whole - rem);
}
}
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