結果
問題 | No.2528 pop_(backfront or not) |
ユーザー |
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提出日時 | 2023-11-03 21:39:27 |
言語 | C++17 (gcc 13.3.0 + boost 1.87.0) |
結果 |
AC
|
実行時間 | 876 ms / 2,000 ms |
コード長 | 2,606 bytes |
コンパイル時間 | 3,807 ms |
コンパイル使用メモリ | 259,860 KB |
最終ジャッジ日時 | 2025-02-17 17:54:03 |
ジャッジサーバーID (参考情報) |
judge4 / judge2 |
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ファイルパターン | 結果 |
---|---|
other | AC * 19 |
ソースコード
#include <bits/stdc++.h>using namespace std;#include <atcoder/all>using namespace atcoder;using ll = long long;using VI = vector<int>;using VVI = vector<VI>;using VL = vector<ll>;using VVL = vector<VL>;using VD = vector<double>;using VVD = vector<VD>;using VS = vector<string>;using P = pair<ll,ll>;using VP = vector<P>;#define rep(i, n) for (ll i = 0; i < ll(n); i++)#define out(x) cout << x << endl#define dout(x) cout << fixed << setprecision(10) << x << endl#define all(a) (a).begin(),(a).end()#define rall(a) (a).rbegin(),(a).rend()#define sz(x) (int)(x.size())#define re0 return 0#define pcnt __builtin_popcountlltemplate<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return 1; } return 0; }template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return 1; } return 0; }constexpr int inf = 1e9;constexpr ll INF = 1e18;//using mint = modint1000000007;using mint = modint998244353;int di[4] = {1,0,-1,0};int dj[4] = {0,1,0,-1};// combination mod prime// https://youtu.be/8uowVvQ_-Mo?t=6002// https://youtu.be/Tgd_zLfRZOQ?t=9928struct 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(){ll n;cin >> n;unordered_map<ll,mint> mem;auto dp = [&](auto &dp,ll lc,ll rc) -> mint {if(mem.count(lc*(2*n+1)+rc)){return mem[lc*(2*n+1)+rc];}if(lc == 1 && rc == 1) return 1;if(lc == 0 || rc == 0) return 0;mint res = 0;res += dp(dp,lc-1,rc-1);if(2 <= lc && 2 <= rc){res += dp(dp,lc-1,rc-1)*(lc-1)*(rc-1);}if(3 <= lc){res += dp(dp,lc-2,rc)*comb(lc-1,2);}if(3 <= rc){res += dp(dp,lc,rc-2)*comb(rc-1,2);}return mem[lc*(2*n+1)+rc] = res;};rep(i,2*n+1){out(dp(dp,i,2*n-i).val());}}