結果
問題 | No.2131 Concon Substrings (COuNt Version) |
ユーザー | stoq |
提出日時 | 2023-02-04 16:06:47 |
言語 | C++17 (gcc 12.3.0 + boost 1.83.0) |
結果 |
AC
|
実行時間 | 96 ms / 2,000 ms |
コード長 | 5,708 bytes |
コンパイル時間 | 4,368 ms |
コンパイル使用メモリ | 270,876 KB |
実行使用メモリ | 113,052 KB |
最終ジャッジ日時 | 2024-07-03 13:19:08 |
合計ジャッジ時間 | 6,375 ms |
ジャッジサーバーID (参考情報) |
judge5 / judge1 |
(要ログイン)
テストケース
テストケース表示入力 | 結果 | 実行時間 実行使用メモリ |
---|---|---|
testcase_00 | AC | 44 ms
112,876 KB |
testcase_01 | AC | 41 ms
113,052 KB |
testcase_02 | AC | 39 ms
113,044 KB |
testcase_03 | AC | 65 ms
112,876 KB |
testcase_04 | AC | 40 ms
112,932 KB |
testcase_05 | AC | 40 ms
112,932 KB |
testcase_06 | AC | 39 ms
112,880 KB |
testcase_07 | AC | 96 ms
112,984 KB |
testcase_08 | AC | 55 ms
112,972 KB |
testcase_09 | AC | 95 ms
112,932 KB |
testcase_10 | AC | 96 ms
112,868 KB |
testcase_11 | AC | 79 ms
112,960 KB |
testcase_12 | AC | 77 ms
112,880 KB |
testcase_13 | AC | 40 ms
112,984 KB |
testcase_14 | AC | 57 ms
113,032 KB |
testcase_15 | AC | 54 ms
112,876 KB |
testcase_16 | AC | 40 ms
112,892 KB |
testcase_17 | AC | 52 ms
112,868 KB |
testcase_18 | AC | 52 ms
112,900 KB |
testcase_19 | AC | 46 ms
113,020 KB |
ソースコード
#define MOD_TYPE 2 #include <bits/stdc++.h> using namespace std; #include <atcoder/all> // #include <atcoder/lazysegtree> // #include <atcoder/modint> // #include <atcoder/segtree> using namespace atcoder; #if 0 #include <boost/multiprecision/cpp_dec_float.hpp> #include <boost/multiprecision/cpp_int.hpp> using Int = boost::multiprecision::cpp_int; using lld = boost::multiprecision::cpp_dec_float_100; #endif #if 0 #include <ext/pb_ds/assoc_container.hpp> #include <ext/pb_ds/tag_and_trait.hpp> #include <ext/pb_ds/tree_policy.hpp> #include <ext/rope> using namespace __gnu_pbds; using namespace __gnu_cxx; template <typename T> using extset = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>; #endif #if 1 #pragma GCC target("avx2") #pragma GCC optimize("O3") #pragma GCC optimize("unroll-loops") #endif #pragma region Macros using ll = long long int; using ld = long double; using pii = pair<int, int>; using pll = pair<ll, ll>; using pld = pair<ld, ld>; template <typename Q_type> using smaller_queue = priority_queue<Q_type, vector<Q_type>, greater<Q_type>>; #if MOD_TYPE == 1 constexpr ll MOD = ll(1e9 + 7); #else #if MOD_TYPE == 2 constexpr ll MOD = 998244353; #else constexpr ll MOD = 1000003; #endif #endif using mint = static_modint<MOD>; constexpr int INF = (int)1e9 + 10; constexpr ll LINF = (ll)4e18; const double PI = acos(-1.0); constexpr ld EPS = 1e-10; constexpr int Dx[] = {0, 0, -1, 1, -1, 1, -1, 1, 0}; constexpr int Dy[] = {1, -1, 0, 0, -1, -1, 1, 1, 0}; #define REP(i, m, n) for (ll i = m; i < (ll)(n); ++i) #define rep(i, n) REP(i, 0, n) #define REPI(i, m, n) for (int i = m; i < (int)(n); ++i) #define repi(i, n) REPI(i, 0, n) #define YES(n) cout << ((n) ? "YES" : "NO") << "\n" #define Yes(n) cout << ((n) ? "Yes" : "No") << "\n" #define all(v) v.begin(), v.end() #define NP(v) next_permutation(all(v)) #define dbg(x) cerr << #x << ":" << x << "\n"; #define UNIQUE(v) v.erase(unique(all(v)), v.end()) struct io_init { io_init() { cin.tie(nullptr); ios::sync_with_stdio(false); cout << setprecision(30) << setiosflags(ios::fixed); }; } io_init; template <typename T> inline bool chmin(T &a, T b) { if (a > b) { a = b; return true; } return false; } template <typename T> inline bool chmax(T &a, T b) { if (a < b) { a = b; return true; } return false; } inline ll floor(ll a, ll b) { if (b < 0) a *= -1, b *= -1; if (a >= 0) return a / b; return -((-a + b - 1) / b); } inline ll ceil(ll a, ll b) { return floor(a + b - 1, b); } template <typename A, size_t N, typename T> inline void Fill(A (&array)[N], const T &val) { fill((T *)array, (T *)(array + N), val); } template <typename T> vector<T> compress(vector<T> &v) { vector<T> val = v; sort(all(val)), val.erase(unique(all(val)), val.end()); for (auto &&vi : v) vi = lower_bound(all(val), vi) - val.begin(); return val; } template <typename T, typename U> constexpr istream &operator>>(istream &is, pair<T, U> &p) noexcept { is >> p.first >> p.second; return is; } template <typename T, typename U> constexpr ostream &operator<<(ostream &os, pair<T, U> p) noexcept { os << p.first << " " << p.second; return os; } ostream &operator<<(ostream &os, mint m) { os << m.val(); return os; } ostream &operator<<(ostream &os, modint m) { os << m.val(); return os; } template <typename T> constexpr istream &operator>>(istream &is, vector<T> &v) noexcept { for (int i = 0; i < v.size(); i++) is >> v[i]; return is; } template <typename T> constexpr ostream &operator<<(ostream &os, vector<T> &v) noexcept { for (int i = 0; i < v.size(); i++) os << v[i] << (i + 1 == v.size() ? "" : " "); return os; } template <typename T> constexpr void operator--(vector<T> &v, int) noexcept { for (int i = 0; i < v.size(); i++) v[i]--; } random_device seed_gen; mt19937_64 engine(seed_gen()); inline ll randInt(ll l, ll r) { return engine() % (r - l + 1) + l; } struct BiCoef { vector<mint> fact_, inv_, finv_; BiCoef(int n) noexcept : fact_(n, 1), inv_(n, 1), finv_(n, 1) { fact_.assign(n, 1), inv_.assign(n, 1), finv_.assign(n, 1); for (int i = 2; i < n; i++) { fact_[i] = fact_[i - 1] * i; inv_[i] = -inv_[MOD % i] * (MOD / i); finv_[i] = finv_[i - 1] * inv_[i]; } } mint C(ll n, ll k) const noexcept { if (n < k || n < 0 || k < 0) return 0; return fact_[n] * finv_[k] * finv_[n - k]; } mint P(ll n, ll k) const noexcept { return C(n, k) * fact_[k]; } mint H(ll n, ll k) const noexcept { return C(n + k - 1, k); } mint Ch1(ll n, ll k) const noexcept { if (n < 0 || k < 0) return 0; mint res = 0; for (int i = 0; i < n; i++) res += C(n, i) * mint(n - i).pow(k) * (i & 1 ? -1 : 1); return res; } mint fact(ll n) const noexcept { if (n < 0) return 0; return fact_[n]; } mint inv(ll n) const noexcept { if (n < 0) return 0; return inv_[n]; } mint finv(ll n) const noexcept { if (n < 0) return 0; return finv_[n]; } }; BiCoef bc(300010); #pragma endregion // ------------------------------- mint dp[3010][3010][3]; void solve() { int n; cin >> n; dp[0][0][0] = 1; rep(i, n) rep(j, i + 1) rep(k, 3) { // c if (k == 0) dp[i + 1][j][1] += dp[i][j][k]; else dp[i + 1][j][k] += dp[i][j][k]; // o if (k == 1) dp[i + 1][j][2] += dp[i][j][k]; else dp[i + 1][j][k] += dp[i][j][k]; // n if (k == 2) dp[i + 1][j + 1][0] += dp[i][j][k]; else dp[i + 1][j][k] += dp[i][j][k]; // else dp[i + 1][j][k] += dp[i][j][k] * 23; } mint ans = 0; rep(j, n + 1) rep(k, 3) ans += dp[n][j][k] * j; cout << ans << "\n"; } int main() { solve(); }