結果
問題 | No.2391 SAN 値チェック |
ユーザー |
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提出日時 | 2023-07-21 22:45:21 |
言語 | C++23 (gcc 13.3.0 + boost 1.87.0) |
結果 |
AC
|
実行時間 | 36 ms / 2,000 ms |
コード長 | 5,297 bytes |
コンパイル時間 | 3,012 ms |
コンパイル使用メモリ | 250,036 KB |
実行使用メモリ | 5,376 KB |
最終ジャッジ日時 | 2024-09-22 00:15:56 |
合計ジャッジ時間 | 4,990 ms |
ジャッジサーバーID (参考情報) |
judge1 / judge3 |
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ファイルパターン | 結果 |
---|---|
other | AC * 17 |
ソースコード
#include <bits/stdc++.h>using namespace std;#define FOR(i,m,n) for(int i=(m);i<(n);++i)#define REP(i,n) FOR(i,0,n)#define ALL(v) (v).begin(),(v).end()using ll = long long;constexpr int INF = 0x3f3f3f3f;constexpr long long LINF = 0x3f3f3f3f3f3f3f3fLL;constexpr double EPS = 1e-8;constexpr int MOD = 998244353;// constexpr int MOD = 1000000007;constexpr int DY4[]{1, 0, -1, 0}, DX4[]{0, -1, 0, 1};constexpr int DY8[]{1, 1, 0, -1, -1, -1, 0, 1};constexpr int DX8[]{0, -1, -1, -1, 0, 1, 1, 1};template <typename T, typename U>inline bool chmax(T& a, U b) { return a < b ? (a = b, true) : false; }template <typename T, typename U>inline bool chmin(T& a, U b) { return a > b ? (a = b, true) : false; }struct IOSetup {IOSetup() {std::cin.tie(nullptr);std::ios_base::sync_with_stdio(false);std::cout << fixed << setprecision(20);}} iosetup;template <unsigned int M>struct MInt {unsigned int v;constexpr MInt() : v(0) {}constexpr MInt(const long long x) : v(x >= 0 ? x % M : x % M + M) {}static constexpr MInt raw(const int x) {MInt x_;x_.v = x;return x_;}static constexpr int get_mod() { return M; }static constexpr void set_mod(const int divisor) {assert(std::cmp_equal(divisor, M));}static void init(const int x) {inv<true>(x);fact(x);fact_inv(x);}template <bool MEMOIZES = false>static MInt inv(const int n) {// assert(0 <= n && n < M && std::gcd(n, M) == 1);static std::vector<MInt> inverse{0, 1};const int prev = inverse.size();if (n < prev) return inverse[n];if constexpr (MEMOIZES) {// "n!" and "M" must be disjoint.inverse.resize(n + 1);for (int i = prev; i <= n; ++i) {inverse[i] = -inverse[M % i] * raw(M / i);}return inverse[n];}int u = 1, v = 0;for (unsigned int a = n, b = M; b;) {const unsigned int q = a / b;std::swap(a -= q * b, b);std::swap(u -= q * v, v);}return u;}static MInt fact(const int n) {static std::vector<MInt> factorial{1};if (const int prev = factorial.size(); n >= prev) {factorial.resize(n + 1);for (int i = prev; i <= n; ++i) {factorial[i] = factorial[i - 1] * i;}}return factorial[n];}static MInt fact_inv(const int n) {static std::vector<MInt> f_inv{1};if (const int prev = f_inv.size(); n >= prev) {f_inv.resize(n + 1);f_inv[n] = inv(fact(n).v);for (int i = n; i > prev; --i) {f_inv[i - 1] = f_inv[i] * i;}}return f_inv[n];}static MInt nCk(const int n, const int k) {if (n < 0 || n < k || k < 0) [[unlikely]] return MInt();return fact(n) * (n - k < k ? fact_inv(k) * fact_inv(n - k) :fact_inv(n - k) * fact_inv(k));}static MInt nPk(const int n, const int k) {return n < 0 || n < k || k < 0 ? MInt() : fact(n) * fact_inv(n - k);}static MInt nHk(const int n, const int k) {return n < 0 || k < 0 ? MInt() : (k == 0 ? 1 : nCk(n + k - 1, k));}static MInt large_nCk(long long n, const int k) {if (n < 0 || n < k || k < 0) [[unlikely]] return MInt();inv<true>(k);MInt res = 1;for (int i = 1; i <= k; ++i) {res *= inv(i) * n--;}return res;}constexpr MInt pow(long long exponent) const {MInt res = 1, tmp = *this;for (; exponent > 0; exponent >>= 1) {if (exponent & 1) res *= tmp;tmp *= tmp;}return res;}constexpr MInt& operator+=(const MInt& x) {if ((v += x.v) >= M) v -= M;return *this;}constexpr MInt& operator-=(const MInt& x) {if ((v += M - x.v) >= M) v -= M;return *this;}constexpr MInt& operator*=(const MInt& x) {v = (unsigned long long){v} * x.v % M;return *this;}MInt& operator/=(const MInt& x) { return *this *= inv(x.v); }constexpr auto operator<=>(const MInt& x) const = default;constexpr MInt& operator++() {if (++v == M) [[unlikely]] v = 0;return *this;}constexpr MInt operator++(int) {const MInt res = *this;++*this;return res;}constexpr MInt& operator--() {v = (v == 0 ? M - 1 : v - 1);return *this;}constexpr MInt operator--(int) {const MInt res = *this;--*this;return res;}constexpr MInt operator+() const { return *this; }constexpr MInt operator-() const { return raw(v ? M - v : 0); }constexpr MInt operator+(const MInt& x) const { return MInt(*this) += x; }constexpr MInt operator-(const MInt& x) const { return MInt(*this) -= x; }constexpr MInt operator*(const MInt& x) const { return MInt(*this) *= x; }MInt operator/(const MInt& x) const { return MInt(*this) /= x; }friend std::ostream& operator<<(std::ostream& os, const MInt& x) {return os << x.v;}friend std::istream& operator>>(std::istream& is, MInt& x) {long long v;is >> v;x = MInt(v);return is;}};using ModInt = MInt<MOD>;// https://math.stackexchange.com/questions/844306/how-to-prove-lim-n-to-infty-en-cdot-left-sum-k-0n-1-k-n-overint main() {int n; cin >> n;cout << 0 << '\n';FOR(k, 1, n + 1) cout << ModInt(-k).pow(n - k) * ModInt::fact_inv(n - k) << '\n';return 0;}