結果
問題 | No.2250 Split Permutation |
ユーザー |
|
提出日時 | 2023-03-13 15:21:43 |
言語 | C++17 (gcc 13.3.0 + boost 1.87.0) |
結果 |
AC
|
実行時間 | 476 ms / 3,000 ms |
コード長 | 2,246 bytes |
コンパイル時間 | 2,118 ms |
コンパイル使用メモリ | 200,696 KB |
最終ジャッジ日時 | 2025-02-11 10:53:07 |
ジャッジサーバーID (参考情報) |
judge2 / judge4 |
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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
#include <bits/stdc++.h>// #include <atcoder/modint>#define rng(a) a.begin(),a.end()#define rrng(a) a.rbegin(),a.rend()#define INF 2000000000000000000#define ll long long#define ull unsigned long long#define ld long double#define pll pair<ll, ll>using namespace std;template<class T>bool chmax(T &a, const T &b) { if (a<b) { a=b; return 1; } return 0; }template<class T>bool chmin(T &a, const T &b) { if (b<a) { a=b; return 1; } return 0; }const double PI = 3.141592653589793;ll mod = 998244353;//=============segment_tree============================struct segment_tree {int MAX_N;int nn;vector<ll> dat;segment_tree(ll n_) {nn = 1;while (nn < n_) {nn *= 2;}MAX_N = nn;dat = vector<ll>(2 * MAX_N - 1, 0);}void update(ll k, ll a) {k += nn - 1;dat.at(k) = a % mod;//要変更while (k > 0) {k = (k - 1) / 2;dat.at(k) = (dat.at(k * 2 + 1) + dat.at(k * 2 + 2)) % mod;//要変更}}ll query(ll a, ll b, ll k = -1, ll l = -1, ll r = -1) {if (k == -1 && l == -1 && r == -1) {k = 0, l = 0, r = nn;}if (r <= a || b <= l) {return 0;//要変更}if (a <= l && r <= b) {return dat.at(k) % mod;}else {ll vl = query(a, b, k * 2 + 1, l, (l + r) / 2);ll vr = query(a, b, k * 2 + 2, (l + r) / 2, r);return (vl + vr) % mod;//要変更}}};//=================================================int main() {ios::sync_with_stdio(false);cin.tie(nullptr);ll N;cin >> N;vector<ll> P(N);for (ll i = 0; i < N; ++i) {cin >> P.at(i);P.at(i) -= 1;}vector<ll> power(N + 1, 1);for (ll i = 0; i < N; ++i) {power.at(i + 1) = power.at(i) * 2 % mod;}ll ans = 0;segment_tree seg1(N);for (ll i = 0; i < N; ++i) {ll p = P.at(i);ll big = seg1.query(p, N);ans += big;ans %= mod;seg1.update(p, 1);}ans *= power.at(N - 1);ans %= mod;segment_tree seg2(N);for (ll i = N - 1; i >= 0; --i) {ll p = P.at(i);// cout << p << "\n";ll small = seg2.query(0, p);ans -= small * power.at(i) % mod;ans = (ans + mod) % mod;seg2.update(p, power.at(N - 1 - i));}cout << ans << "\n";}