結果
問題 | No.2791 Beginner Contest |
ユーザー | 👑 binap |
提出日時 | 2024-06-21 21:49:46 |
言語 | C++17(gcc12) (gcc 12.3.0 + boost 1.87.0) |
結果 |
AC
|
実行時間 | 68 ms / 2,000 ms |
コード長 | 2,642 bytes |
コンパイル時間 | 4,956 ms |
コンパイル使用メモリ | 267,128 KB |
実行使用メモリ | 6,944 KB |
最終ジャッジ日時 | 2024-06-21 21:49:52 |
合計ジャッジ時間 | 6,142 ms |
ジャッジサーバーID (参考情報) |
judge5 / judge2 |
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ファイルパターン | 結果 |
---|---|
sample | AC * 3 |
other | AC * 17 |
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; 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 <int m> ostream& operator<<(ostream& os, const dynamic_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; 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; using S = mint; S op(S a, S b){ return a + b; } S e(){ return (S)0; } using F = mint; S mapping(F f, S x){ return f + x; } F composition(F f, F g){ return f + g; } F id(){ return (S)0; } int main(){ int n, k; cin >> n >> k; lazy_segtree<S, op, e, F, mapping, composition, id> seg(n + 1); seg.set(0, (S)1); for(int from = 0; from <= n - k; from++){ seg.apply(from + k, n + 1, seg.get(from)); } S ans = 0; rep(i, n + 1) ans += seg.get(i); cout << ans << "\n"; return 0; }