結果
問題 | No.1296 OR or NOR |
ユーザー | koba-e964 |
提出日時 | 2021-09-16 02:06:35 |
言語 | Rust (1.77.0 + proconio) |
結果 |
RE
|
実行時間 | - |
コード長 | 4,840 bytes |
コンパイル時間 | 16,337 ms |
コンパイル使用メモリ | 401,016 KB |
実行使用メモリ | 8,892 KB |
最終ジャッジ日時 | 2024-06-29 03:02:06 |
合計ジャッジ時間 | 23,009 ms |
ジャッジサーバーID (参考情報) |
judge5 / judge3 |
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テストケース
テストケース表示入力 | 結果 | 実行時間 実行使用メモリ |
---|---|---|
testcase_00 | RE | - |
testcase_01 | AC | 2 ms
5,376 KB |
testcase_02 | RE | - |
testcase_03 | RE | - |
testcase_04 | RE | - |
testcase_05 | RE | - |
testcase_06 | RE | - |
testcase_07 | RE | - |
testcase_08 | RE | - |
testcase_09 | RE | - |
testcase_10 | RE | - |
testcase_11 | RE | - |
testcase_12 | RE | - |
testcase_13 | RE | - |
testcase_14 | RE | - |
testcase_15 | RE | - |
testcase_16 | RE | - |
testcase_17 | AC | 99 ms
8,320 KB |
testcase_18 | AC | 107 ms
8,320 KB |
testcase_19 | AC | 97 ms
8,320 KB |
testcase_20 | RE | - |
testcase_21 | RE | - |
testcase_22 | RE | - |
testcase_23 | RE | - |
testcase_24 | RE | - |
testcase_25 | RE | - |
testcase_26 | RE | - |
testcase_27 | RE | - |
testcase_28 | RE | - |
testcase_29 | RE | - |
testcase_30 | RE | - |
testcase_31 | RE | - |
testcase_32 | RE | - |
testcase_33 | RE | - |
testcase_34 | RE | - |
コンパイルメッセージ
warning: unused variable: `whole` --> src/main.rs:105:9 | 105 | let whole = (1i64 << 60) - 1; | ^^^^^ help: if this is intentional, prefix it with an underscore: `_whole` | = note: `#[warn(unused_variables)]` on by default
ソースコード
#[allow(unused_imports)] use std::cmp::*; #[allow(unused_imports)] use std::collections::*; 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::<$t>().expect("Parse error")); } /** * Segment Tree. This data structure is useful for fast folding on intervals of an array * whose elements are elements of monoid I. Note that constructing this tree requires the identity * element of I and the operation of I. * Verified by: yukicoder No. 259 (http://yukicoder.me/submissions/100581) * AGC015-E (http://agc015.contest.atcoder.jp/submissions/1461001) */ struct SegTree<I, BiOp> { n: usize, dat: Vec<I>, op: BiOp, e: I, } impl<I, BiOp> SegTree<I, BiOp> where BiOp: Fn(I, I) -> I, I: Copy { pub fn new(n_: usize, op: BiOp, e: I) -> Self { let mut n = 1; while n < n_ { n *= 2; } // n is a power of 2 SegTree {n: n, dat: vec![e; 2 * n - 1], op: op, e: e} } /* ary[k] <- v */ pub fn update(&mut self, idx: usize, v: I) { let mut k = idx + self.n - 1; self.dat[k] = v; while k > 0 { k = (k - 1) / 2; self.dat[k] = (self.op)(self.dat[2 * k + 1], self.dat[2 * k + 2]); } } /* [a, b) (note: half-inclusive) * http://proc-cpuinfo.fixstars.com/2017/07/optimize-segment-tree/ */ pub fn query(&self, mut a: usize, mut b: usize) -> I { let mut left = self.e; let mut right = self.e; a += self.n - 1; b += self.n - 1; while a < b { if (a & 1) == 0 { left = (self.op)(left, self.dat[a]); } if (b & 1) == 0 { right = (self.op)(self.dat[b - 1], right); } a = a / 2; b = (b - 1) / 2; } (self.op)(left, right) } } fn solve() { let out = std::io::stdout(); let mut out = BufWriter::new(out.lock()); macro_rules! puts { ($($format:tt)*) => (let _ = write!(out,$($format)*);); } input! { n: usize, a: [i64; n], q: usize, b: [i64; q], } let mut st = SegTree::new(n, |x, y| x | y, 0i64); for i in 0..n { st.update(i, a[i]); } let whole = (1i64 << 60) - 1; let whole = 1023; for b in b { let mut targ = b; let mut rem = whole; let mut cur = n; let mut ok = true; let mut ans = 0; let last = a[n - 1]; if (last & b) != last { targ ^= rem; ans += 1; if (targ & last) != last { ok = false; } } while cur > 0 && rem != 0 && ok { assert_eq!(rem & targ, targ); let mut fail = 0; let mut pass = cur + 1; while pass - fail > 1 { let mid = (pass + fail) / 2; let o = st.query(mid - 1, cur); if ((o & rem) | targ) == targ { pass = mid; } else { fail = mid; } } if pass == cur + 1 { ok = false; break; } let o = st.query(pass - 1, cur); // eprintln!("b = {}, [{}, {}) o = {} ({} <= {})", b, pass - 1, cur, o, targ, rem); rem &= !o; targ &= rem; cur = pass - 1; if cur != 0 { targ ^= rem; ans += 1; } } // TODO: prove it's ok to flip "before the first element" ok &= targ == 0 || targ == rem; if targ == rem { ans += 1; } puts!("{}\n", if ok { ans } else { -1 }); } } 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(); }