結果
問題 | No.2250 Split Permutation |
ユーザー |
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提出日時 | 2023-03-17 23:12:05 |
言語 | C++17 (gcc 13.3.0 + boost 1.87.0) |
結果 |
AC
|
実行時間 | 74 ms / 3,000 ms |
コード長 | 6,793 bytes |
コンパイル時間 | 3,945 ms |
コンパイル使用メモリ | 261,296 KB |
最終ジャッジ日時 | 2025-02-11 14:16:17 |
ジャッジサーバーID (参考情報) |
judge3 / judge2 |
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ファイルパターン | 結果 |
---|---|
sample | AC * 3 |
other | AC * 35 |
ソースコード
#define MOD_TYPE 2#include <bits/stdc++.h>using namespace std;#include <atcoder/all>// #include <atcoder/lazysegtree>// #include <atcoder/modint>// #include <atcoder/segtree>using namespace atcoder;#if 0#include <boost/multiprecision/cpp_dec_float.hpp>#include <boost/multiprecision/cpp_int.hpp>using Int = boost::multiprecision::cpp_int;using lld = boost::multiprecision::cpp_dec_float_100;#endif#if 0#include <ext/pb_ds/assoc_container.hpp>#include <ext/pb_ds/tag_and_trait.hpp>#include <ext/pb_ds/tree_policy.hpp>#include <ext/rope>using namespace __gnu_pbds;using namespace __gnu_cxx;template <typename T>using extset =tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;#endif#if 0#pragma GCC target("avx2")#pragma GCC optimize("O3")#pragma GCC optimize("unroll-loops")#endif#pragma region Macrosusing ll = long long int;using ld = long double;using pii = pair<int, int>;using pll = pair<ll, ll>;using pld = pair<ld, ld>;template <typename Q_type>using smaller_queue = priority_queue<Q_type, vector<Q_type>, greater<Q_type>>;#if MOD_TYPE == 1constexpr ll MOD = ll(1e9 + 7);#else#if MOD_TYPE == 2constexpr ll MOD = 998244353;#elseconstexpr ll MOD = 1000003;#endif#endifusing mint = static_modint<MOD>;constexpr int INF = (int)1e9 + 10;constexpr ll LINF = (ll)4e18;const double PI = acos(-1.0);constexpr ld EPS = 1e-10;constexpr int Dx[] = {0, 0, -1, 1, -1, 1, -1, 1, 0};constexpr int Dy[] = {1, -1, 0, 0, -1, -1, 1, 1, 0};#define REP(i, m, n) for (ll i = m; i < (ll)(n); ++i)#define rep(i, n) REP(i, 0, n)#define REPI(i, m, n) for (int i = m; i < (int)(n); ++i)#define repi(i, n) REPI(i, 0, n)#define RREP(i, m, n) for (ll i = n - 1; i >= m; i--)#define rrep(i, n) RREP(i, 0, n)#define YES(n) cout << ((n) ? "YES" : "NO") << "\n"#define Yes(n) cout << ((n) ? "Yes" : "No") << "\n"#define all(v) v.begin(), v.end()#define NP(v) next_permutation(all(v))#define dbg(x) cerr << #x << ":" << x << "\n";#define UNIQUE(v) v.erase(unique(all(v)), v.end())struct io_init {io_init() {cin.tie(nullptr);ios::sync_with_stdio(false);cout << setprecision(30) << setiosflags(ios::fixed);};} io_init;template <typename T>inline bool chmin(T &a, T b) {if (a > b) {a = b;return true;}return false;}template <typename T>inline bool chmax(T &a, T b) {if (a < b) {a = b;return true;}return false;}inline ll floor(ll a, ll b) {if (b < 0) a *= -1, b *= -1;if (a >= 0) return a / b;return -((-a + b - 1) / b);}inline ll ceil(ll a, ll b) { return floor(a + b - 1, b); }template <typename A, size_t N, typename T>inline void Fill(A (&array)[N], const T &val) {fill((T *)array, (T *)(array + N), val);}template <typename T>vector<T> compress(vector<T> &v) {vector<T> val = v;sort(all(val)), val.erase(unique(all(val)), val.end());for (auto &&vi : v) vi = lower_bound(all(val), vi) - val.begin();return val;}template <typename T, typename U>constexpr istream &operator>>(istream &is, pair<T, U> &p) noexcept {is >> p.first >> p.second;return is;}template <typename T, typename U>constexpr ostream &operator<<(ostream &os, pair<T, U> p) noexcept {os << p.first << " " << p.second;return os;}ostream &operator<<(ostream &os, mint m) {os << m.val();return os;}ostream &operator<<(ostream &os, modint m) {os << m.val();return os;}template <typename T>constexpr istream &operator>>(istream &is, vector<T> &v) noexcept {for (int i = 0; i < v.size(); i++) is >> v[i];return is;}template <typename T>constexpr ostream &operator<<(ostream &os, vector<T> &v) noexcept {for (int i = 0; i < v.size(); i++)os << v[i] << (i + 1 == v.size() ? "" : " ");return os;}template <typename T>constexpr void operator--(vector<T> &v, int) noexcept {for (int i = 0; i < v.size(); i++) v[i]--;}random_device seed_gen;mt19937_64 engine(seed_gen());inline ll randInt(ll l, ll r) { return engine() % (r - l + 1) + l; }struct BiCoef {vector<mint> fact_, inv_, finv_;BiCoef(int n) noexcept : fact_(n, 1), inv_(n, 1), finv_(n, 1) {fact_.assign(n, 1), inv_.assign(n, 1), finv_.assign(n, 1);for (int i = 2; i < n; i++) {fact_[i] = fact_[i - 1] * i;inv_[i] = -inv_[MOD % i] * (MOD / i);finv_[i] = finv_[i - 1] * inv_[i];}}mint C(ll n, ll k) const noexcept {if (n < k || n < 0 || k < 0) return 0;return fact_[n] * finv_[k] * finv_[n - k];}mint P(ll n, ll k) const noexcept { return C(n, k) * fact_[k]; }mint H(ll n, ll k) const noexcept { return C(n + k - 1, k); }mint Ch1(ll n, ll k) const noexcept {if (n < 0 || k < 0) return 0;mint res = 0;for (int i = 0; i < n; i++)res += C(n, i) * mint(n - i).pow(k) * (i & 1 ? -1 : 1);return res;}mint fact(ll n) const noexcept {if (n < 0) return 0;return fact_[n];}mint inv(ll n) const noexcept {if (n < 0) return 0;return inv_[n];}mint finv(ll n) const noexcept {if (n < 0) return 0;return finv_[n];}};BiCoef bc(100010);#pragma endregion// -------------------------------template <typename T>struct BIT {int n;vector<T> dat;BIT() {}BIT(int n) : n(n), dat(n + 1, 0) {}BIT(vector<T> &a) : n(a.size()), dat(a.size() + 1) {for (int i = 0; i < n; i++) dat[i + 1] = a[i];for (int i = 1; i <= n; i++) {if (i + (i & -i) <= n) dat[i + (i & -i)] += dat[i];}}// 0-indexedvoid add(int i, T x) {i++;while (i <= n) {dat[i] += x;i += i & -i;}}void update(int i, T x) { add(i, x - get(i)); }// [0, i)T sum(int i) {T res = 0;while (i > 0) {res += dat[i];i -= i & -i;}return res;}// 0-indexedT get(int i) { return sum(i + 1) - sum(i); }// [l, r)T query(int l, int r) { return sum(r) - sum(l); }void display() {for (int i = 0; i < n; i++) cerr << get(i) << " ";cerr << "\n";}};template <typename T>ll inversion(vector<T> a, bool zaatsu = false) {int z;if (zaatsu) {map<T, int> mp;z = 0;for (auto &&ai : a) mp[ai] = 0;for (auto &&[k, v] : mp) v = z++;for (auto &&ai : a) ai = mp[ai];} else {z = a.size();}BIT<ll> bit(z);ll res = 0;for (int i = 0; i < a.size(); i++) {res += bit.query(a[i] + 1, z);bit.add(a[i], 1);}return res;}void solve() {int n;cin >> n;vector<int> p(n);cin >> p;p--;mint inv = inversion(p);BIT<mint> bit(n);mint t = 0;rep(i, n) {mint sum = bit.query(p[i] + 1, n) * bc.inv(2).pow(i);t += sum;bit.add(p[i], mint(2).pow(i));}mint ans = (inv - t) * mint(2).pow(n - 1);cout << ans << "\n";}int main() { solve(); }