結果
問題 |
No.3043 括弧列の数え上げ
|
ユーザー |
👑 |
提出日時 | 2025-02-28 22:38:27 |
言語 | C++17 (gcc 13.3.0 + boost 1.87.0) |
結果 |
AC
|
実行時間 | 6 ms / 2,000 ms |
コード長 | 3,633 bytes |
コンパイル時間 | 4,314 ms |
コンパイル使用メモリ | 255,252 KB |
実行使用メモリ | 6,820 KB |
最終ジャッジ日時 | 2025-02-28 22:38:33 |
合計ジャッジ時間 | 5,435 ms |
ジャッジサーバーID (参考情報) |
judge1 / judge4 |
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ファイルパターン | 結果 |
---|---|
sample | AC * 3 |
other | AC * 45 |
ソースコード
#include<bits/stdc++.h> #include<atcoder/all> #define rep(i,n) for(int i=0;i<n;i++) using namespace std; using namespace atcoder; typedef long long ll; typedef vector<int> vi; typedef vector<long long> vl; typedef vector<vector<int>> vvi; typedef vector<vector<long long>> vvl; typedef long double ld; typedef pair<int, int> P; template <int m> ostream& operator<<(ostream& os, const static_modint<m>& a) {os << a.val(); return os;} template <int m> ostream& operator<<(ostream& os, const dynamic_modint<m>& a) {os << a.val(); return os;} template <int m> istream& operator>>(istream& is, static_modint<m>& a) {long long x; is >> x; a = x; return is;} template <int m> istream& operator>>(istream& is, dynamic_modint<m>& a) {long long x; is >> x; a = x; return is;} template<typename T> istream& operator>>(istream& is, vector<T>& v){int n = v.size(); assert(n > 0); rep(i, n) is >> v[i]; return is;} template<typename U, typename T> ostream& operator<<(ostream& os, const pair<U, T>& p){os << p.first << ' ' << p.second; return os;} template<typename T> ostream& operator<<(ostream& os, const vector<T>& v){int n = v.size(); rep(i, n) os << v[i] << (i == n - 1 ? "\n" : " "); return os;} template<typename T> ostream& operator<<(ostream& os, const vector<vector<T>>& v){int n = v.size(); rep(i, n) os << v[i] << (i == n - 1 ? "\n" : ""); return os;} template<typename T> ostream& operator<<(ostream& os, const set<T>& se){for(T x : se) os << x << " "; os << "\n"; return os;} template<typename T> ostream& operator<<(ostream& os, const unordered_set<T>& se){for(T x : se) os << x << " "; os << "\n"; return os;} template<typename S, auto op, auto e> ostream& operator<<(ostream& os, const atcoder::segtree<S, op, e>& seg){int n = seg.max_right(0, [](S){return true;}); rep(i, n) os << seg.get(i) << (i == n - 1 ? "\n" : " "); return os;} template<typename S, auto op, auto e, typename F, auto mapping, auto composition, auto id> ostream& operator<<(ostream& os, const atcoder::lazy_segtree<S, op, e, F, mapping, composition, id>& seg){int n = seg.max_right(0, [](S){return true;}); rep(i, n) os << seg.get(i) << (i == n - 1 ? "\n" : " "); return os;} template<typename T> void chmin(T& a, T b){a = min(a, b);} template<typename T> void chmax(T& a, T b){a = max(a, b);} using mint = modint998244353; // combination mod prime // https://youtu.be/8uowVvQ_-Mo?t=6002 // https://youtu.be/Tgd_zLfRZOQ?t=9928 struct modinv { int n; vector<mint> d; modinv(): n(2), d({0,1}) {} mint operator()(int i) { while (n <= i) d.push_back(-d[mint::mod()%n]*(mint::mod()/n)), ++n; return d[i]; } mint operator[](int i) const { return d[i];} } invs; struct modfact { int n; vector<mint> d; modfact(): n(2), d({1,1}) {} mint operator()(int i) { while (n <= i) d.push_back(d.back()*n), ++n; return d[i]; } mint operator[](int i) const { return d[i];} } facts; struct modfactinv { int n; vector<mint> d; modfactinv(): n(2), d({1,1}) {} mint operator()(int i) { while (n <= i) d.push_back(d.back()*invs(n)), ++n; return d[i]; } mint operator[](int i) const { return d[i];} } ifacts; mint comb(int n, int k) { if (n < k || k < 0) return 0; return facts(n)*ifacts(k)*ifacts(n-k); } int main(){ int N; cin >> N; if(N % 2){ cout << "0\n"; return 0; } N /= 2; vector<mint> c(N + 1); vector<mint> ans(N + 1); for(int n = 0; n <= N; n++){ c[n] = comb(2 * n, n) / (n + 1); } for(int n = 0; n <= N; n++){ for(int n1 = 0; n1 <= n - 1; n1++){ int n2 = (n - 1) - n1; ans[n] += c[n1] * c[n2] * n2 + ans[n1] * c[n2] + c[n1] * ans[n2]; } } cout << ans[N] << "\n"; return 0; }