結果

問題 No.2528 pop_(backfront or not)
ユーザー 👑 binap
提出日時 2023-11-03 22:21:04
言語 C++17
(gcc 13.3.0 + boost 1.87.0)
結果
AC  
実行時間 186 ms / 2,000 ms
コード長 2,595 bytes
コンパイル時間 4,592 ms
コンパイル使用メモリ 258,244 KB
最終ジャッジ日時 2025-02-17 18:28:52
ジャッジサーバーID
(参考情報)
judge5 / judge4
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ファイルパターン 結果
other AC * 19
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ソースコード

diff #
プレゼンテーションモードにする

#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;
ostream& operator<<(ostream& os, const modint& a) {os << a.val(); return os;}
template <int m> ostream& operator<<(ostream& os, const static_modint<m>& a) {os << a.val(); return os;}
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 << "\n"; 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;}
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);
}
vector<vector<mint>> memo(2001);
vector<mint> solve(int n){
if(memo[n].size()) return memo[n];
if(n == 1) return memo[1] = {0, 1, 0};
vector<mint> res(2 * n + 1);
auto small = solve(n - 1);
rep(i, 2 * n - 1) res[i + 1] += small[i];
rep(i, 2 * n - 1){
res[i + 1] += comb(i, 2) * small[i - 1];
res[i + 1] += i * (2 * n - i - 2) * small[i];
res[i + 1] += comb(2 * n - i - 2, 2) * small[i + 1];
}
return memo[n] = res;
}
int main(){
int n;
cin >> n;
auto ans = solve(n);
for(auto x : ans){
cout << x << "\n";
}
return 0;
}
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