結果
問題 | No.1435 Mmm...... |
ユーザー |
|
提出日時 | 2021-11-29 22:09:24 |
言語 | Rust (1.83.0 + proconio) |
結果 |
AC
|
実行時間 | 112 ms / 2,000 ms |
コード長 | 5,502 bytes |
コンパイル時間 | 11,290 ms |
コンパイル使用メモリ | 402,332 KB |
実行使用メモリ | 15,848 KB |
最終ジャッジ日時 | 2024-07-02 13:41:21 |
合計ジャッジ時間 | 15,149 ms |
ジャッジサーバーID (参考情報) |
judge3 / judge5 |
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ファイルパターン | 結果 |
---|---|
sample | AC * 4 |
other | AC * 24 |
ソースコード
use std::cmp::*;// https://qiita.com/tanakh/items/0ba42c7ca36cd29d0ac8macro_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 2SegTree {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)}}// Depends on: datastr/SegTree.rs// Verified by: https://yukicoder.me/submissions/717436impl<I, BiOp> SegTree<I, BiOp>where BiOp: Fn(I, I) -> I,I: Copy {// Port from https://github.com/atcoder/ac-library/blob/master/atcoder/segtree.hpp#[allow(unused)]fn max_right<F: Fn(I) -> bool>(&self, mut l: usize, f: &F,) -> usize {assert!(f(self.e));if l == self.orign {return self.orign;}l += self.n - 1;let mut sm = self.e;loop {while l % 2 == 1 {l = (l - 1) / 2;}if !f((self.op)(sm, self.dat[l])) {while l < self.n - 1 {l = 2 * l + 1;let val = (self.op)(sm, self.dat[l]);if f(val) {sm = val;l += 1;}}return std::cmp::min(self.orign, l + 1 - self.n);}sm = (self.op)(sm, self.dat[l]);l += 1;if (l + 1).is_power_of_two() { break; }}self.orign}// Port from https://github.com/atcoder/ac-library/blob/master/atcoder/segtree.hpp#[allow(unused)]fn min_left<F: Fn(I) -> bool>(&self, mut r: usize, f: &F,) -> usize {if !f(self.e) {return r + 1;}if r == 0 {return 0;}r += self.n - 1;let mut sm = self.e;loop {r -= 1;while r > 0 && r % 2 == 0 {r = (r - 1) / 2;}if !f((self.op)(self.dat[r], sm)) {while r < self.n - 1 {r = 2 * r + 2;let val = (self.op)(self.dat[r], sm);if f(val) {sm = val;r -= 1;}}return r + 2 - self.n;}sm = (self.op)(self.dat[r], sm);if (r + 1).is_power_of_two() { break; }}0}}const INF: i64 = 1 << 50;fn main() {input! {n: usize,a: [i64; n],}let mut st = SegTree::new(n, |(a1, a2, b), (c1, c2, d)| {let mut v = [a1, a2, c1, c2];v.sort();(v[0], v[1], max(b, d))}, (INF, INF, -INF));for i in 0..n {st.update(i, (a[i], INF, a[i]));}let mut ans = 0i64;for i in 0..n {let idx = st.max_right(i, &|(m1, m2, ma)| m1 + m2 >= ma);ans += (idx - i - 1) as i64;}println!("{}", ans);}