結果

問題 No.2250 Split Permutation
ユーザー koba-e964koba-e964
提出日時 2023-07-30 01:08:27
言語 Rust
(1.77.0 + proconio)
結果
AC  
実行時間 163 ms / 3,000 ms
コード長 7,304 bytes
コンパイル時間 14,695 ms
コンパイル使用メモリ 405,496 KB
実行使用メモリ 11,648 KB
最終ジャッジ日時 2024-10-08 13:10:29
合計ジャッジ時間 18,611 ms
ジャッジサーバーID
(参考情報)
judge1 / judge2
このコードへのチャレンジ
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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 AC 1 ms
5,248 KB
testcase_01 AC 1 ms
5,248 KB
testcase_02 AC 1 ms
5,248 KB
testcase_03 AC 1 ms
5,248 KB
testcase_04 AC 1 ms
5,248 KB
testcase_05 AC 1 ms
5,248 KB
testcase_06 AC 1 ms
5,248 KB
testcase_07 AC 1 ms
5,248 KB
testcase_08 AC 1 ms
5,248 KB
testcase_09 AC 1 ms
5,248 KB
testcase_10 AC 1 ms
5,248 KB
testcase_11 AC 1 ms
5,248 KB
testcase_12 AC 1 ms
5,248 KB
testcase_13 AC 1 ms
5,248 KB
testcase_14 AC 1 ms
5,248 KB
testcase_15 AC 1 ms
5,248 KB
testcase_16 AC 1 ms
5,248 KB
testcase_17 AC 1 ms
5,248 KB
testcase_18 AC 162 ms
11,648 KB
testcase_19 AC 147 ms
11,512 KB
testcase_20 AC 128 ms
11,264 KB
testcase_21 AC 71 ms
6,912 KB
testcase_22 AC 109 ms
11,136 KB
testcase_23 AC 19 ms
5,248 KB
testcase_24 AC 111 ms
11,136 KB
testcase_25 AC 161 ms
11,648 KB
testcase_26 AC 51 ms
6,528 KB
testcase_27 AC 13 ms
5,248 KB
testcase_28 AC 25 ms
5,248 KB
testcase_29 AC 160 ms
11,648 KB
testcase_30 AC 27 ms
5,248 KB
testcase_31 AC 7 ms
5,248 KB
testcase_32 AC 33 ms
5,248 KB
testcase_33 AC 81 ms
6,912 KB
testcase_34 AC 19 ms
5,248 KB
testcase_35 AC 56 ms
6,656 KB
testcase_36 AC 32 ms
5,248 KB
testcase_37 AC 163 ms
11,516 KB
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ソースコード

diff #

// 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, usize1) => (read_value!($next, usize) - 1);
    ($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. 2220 (https://yukicoder.me/submissions/841554)
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) (half-inclusive)
    // http://proc-cpuinfo.fixstars.com/2017/07/optimize-segment-tree/
    #[allow(unused)]
    pub fn query(&self, rng: std::ops::Range<usize>) -> I {
        let (mut a, mut b) = (rng.start, rng.end);
        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)
    }
}

/// Verified by https://atcoder.jp/contests/abc198/submissions/21774342
mod mod_int {
    use std::ops::*;
    pub trait Mod: Copy { fn m() -> i64; }
    #[derive(Copy, Clone, Hash, PartialEq, Eq, PartialOrd, Ord)]
    pub struct ModInt<M> { pub x: i64, phantom: ::std::marker::PhantomData<M> }
    impl<M: Mod> ModInt<M> {
        // x >= 0
        pub fn new(x: i64) -> Self { ModInt::new_internal(x % M::m()) }
        fn new_internal(x: i64) -> Self {
            ModInt { x: x, phantom: ::std::marker::PhantomData }
        }
        pub fn pow(self, mut e: i64) -> Self {
            debug_assert!(e >= 0);
            let mut sum = ModInt::new_internal(1);
            let mut cur = self;
            while e > 0 {
                if e % 2 != 0 { sum *= cur; }
                cur *= cur;
                e /= 2;
            }
            sum
        }
        #[allow(dead_code)]
        pub fn inv(self) -> Self { self.pow(M::m() - 2) }
    }
    impl<M: Mod> Default for ModInt<M> {
        fn default() -> Self { Self::new_internal(0) }
    }
    impl<M: Mod, T: Into<ModInt<M>>> Add<T> for ModInt<M> {
        type Output = Self;
        fn add(self, other: T) -> Self {
            let other = other.into();
            let mut sum = self.x + other.x;
            if sum >= M::m() { sum -= M::m(); }
            ModInt::new_internal(sum)
        }
    }
    impl<M: Mod, T: Into<ModInt<M>>> Sub<T> for ModInt<M> {
        type Output = Self;
        fn sub(self, other: T) -> Self {
            let other = other.into();
            let mut sum = self.x - other.x;
            if sum < 0 { sum += M::m(); }
            ModInt::new_internal(sum)
        }
    }
    impl<M: Mod, T: Into<ModInt<M>>> Mul<T> for ModInt<M> {
        type Output = Self;
        fn mul(self, other: T) -> Self { ModInt::new(self.x * other.into().x % M::m()) }
    }
    impl<M: Mod, T: Into<ModInt<M>>> AddAssign<T> for ModInt<M> {
        fn add_assign(&mut self, other: T) { *self = *self + other; }
    }
    impl<M: Mod, T: Into<ModInt<M>>> SubAssign<T> for ModInt<M> {
        fn sub_assign(&mut self, other: T) { *self = *self - other; }
    }
    impl<M: Mod, T: Into<ModInt<M>>> MulAssign<T> for ModInt<M> {
        fn mul_assign(&mut self, other: T) { *self = *self * other; }
    }
    impl<M: Mod> Neg for ModInt<M> {
        type Output = Self;
        fn neg(self) -> Self { ModInt::new(0) - self }
    }
    impl<M> ::std::fmt::Display for ModInt<M> {
        fn fmt(&self, f: &mut ::std::fmt::Formatter) -> ::std::fmt::Result {
            self.x.fmt(f)
        }
    }
    impl<M: Mod> ::std::fmt::Debug for ModInt<M> {
        fn fmt(&self, f: &mut ::std::fmt::Formatter) -> ::std::fmt::Result {
            let (mut a, mut b, _) = red(self.x, M::m());
            if b < 0 {
                a = -a;
                b = -b;
            }
            write!(f, "{}/{}", a, b)
        }
    }
    impl<M: Mod> From<i64> for ModInt<M> {
        fn from(x: i64) -> Self { Self::new(x) }
    }
    // Finds the simplest fraction x/y congruent to r mod p.
    // The return value (x, y, z) satisfies x = y * r + z * p.
    fn red(r: i64, p: i64) -> (i64, i64, i64) {
        if r.abs() <= 10000 {
            return (r, 1, 0);
        }
        let mut nxt_r = p % r;
        let mut q = p / r;
        if 2 * nxt_r >= r {
            nxt_r -= r;
            q += 1;
        }
        if 2 * nxt_r <= -r {
            nxt_r += r;
            q -= 1;
        }
        let (x, z, y) = red(nxt_r, r);
        (x, y - q * z, z)
    }
} // mod mod_int

macro_rules! define_mod {
    ($struct_name: ident, $modulo: expr) => {
        #[derive(Copy, Clone, PartialEq, Eq, PartialOrd, Ord, Hash)]
        pub struct $struct_name {}
        impl mod_int::Mod for $struct_name { fn m() -> i64 { $modulo } }
    }
}
const MOD: i64 = 998_244_353;
define_mod!(P, MOD);
type MInt = mod_int::ModInt<P>;

// https://yukicoder.me/problems/no/2250 (3)
// p[i] > p[j] のときの 2^{N-1}(1 - 2^{-i} 2^j) の和を求めれば良い。
fn main() {
    input! {
        n: usize,
        p: [usize1; n],
    }
    let mut st = SegTree::new(n, |x, y| x + y, MInt::new(0));
    let mut st2 = SegTree::new(n, |x, y| x + y, MInt::new(0));
    let mut tot = MInt::new(0);
    for i in 0..n {
        tot += st.query(p[i] + 1..n);
        tot -= st2.query(p[i] + 1..n) * MInt::new(2).pow(MOD - 1 - i as i64);
        st.update(p[i], MInt::new(1));
        st2.update(p[i], MInt::new(2).pow(i as i64));
    }
    println!("{}", tot * MInt::new(2).pow(n as i64 - 1));
}
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