結果
| 問題 |
No.2883 K-powered Sum of Fibonacci
|
| コンテスト | |
| ユーザー |
SnowBeenDiding
|
| 提出日時 | 2024-09-09 03:55:35 |
| 言語 | C++23 (gcc 13.3.0 + boost 1.87.0) |
| 結果 |
AC
|
| 実行時間 | 503 ms / 3,000 ms |
| コード長 | 5,019 bytes |
| コンパイル時間 | 5,734 ms |
| コンパイル使用メモリ | 314,192 KB |
| 実行使用メモリ | 42,564 KB |
| 最終ジャッジ日時 | 2024-09-09 03:55:51 |
| 合計ジャッジ時間 | 15,039 ms |
|
ジャッジサーバーID (参考情報) |
judge3 / judge2 |
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| ファイルパターン | 結果 |
|---|---|
| sample | AC * 3 |
| other | AC * 40 |
ソースコード
#include <atcoder/all>
#include <bits/stdc++.h>
#define rep(i, a, b) for (ll i = (ll)(a); i < (ll)(b); i++)
using namespace atcoder;
using namespace std;
typedef long long ll;
using mint = modint998244353;
struct Comb {
vector<mint> fact, ifact;
int MAX_COM;
Comb() {}
Comb(int n) {
MAX_COM = n; // ここでMAX入力を調整
init(998244353, MAX_COM);
}
void init(long long MOD, long long MAX_COM) {
int n = MAX_COM;
assert(n < MOD);
fact = vector<mint>(n + 1);
ifact = vector<mint>(n + 1);
fact[0] = 1;
for (int i = 1; i <= n; ++i)
fact[i] = fact[i - 1] * i;
ifact[n] = fact[n].inv();
for (int i = n; i >= 1; --i)
ifact[i - 1] = ifact[i] * i;
}
mint operator()(long long n, long long k) {
if (k < 0 || k > n)
return 0;
return fact[n] * ifact[k] * ifact[n - k];
}
};
Comb comb(5000010);
template <class T> struct Matrix {
vector<vector<T>> A;
Matrix() {}
Matrix(size_t n, size_t m) : A(n, std::vector<T>(m, zero())) {}
Matrix(size_t n) : A(n, std::vector<T>(n, zero())) {};
T zero() { return (T(0)); }
size_t height() const { return (A.size()); }
size_t width() const { return (A[0].size()); }
inline const vector<T> &operator[](int k) const { return (A.at(k)); }
inline vector<T> &operator[](int k) { return (A.at(k)); }
static Matrix I(size_t n) {
Matrix mat(n);
for (int i = 0; i < n; i++)
mat[i][i] = 1;
return (mat);
}
Matrix &operator+=(const Matrix &B) {
size_t n = height(), m = width();
assert(n == B.height() && m == B.width());
for (int i = 0; i < n; i++)
for (int j = 0; j < m; j++)
(*this)[i][j] += B[i][j];
return (*this);
}
Matrix &operator-=(const Matrix &B) {
size_t n = height(), m = width();
assert(n == B.height() && m == B.width());
for (int i = 0; i < n; i++)
for (int j = 0; j < m; j++)
(*this)[i][j] -= B[i][j];
return (*this);
}
Matrix &operator*=(const Matrix &B) {
size_t n = height(), m = B.width(), p = width();
assert(p == B.height());
vector<vector<T>> C(n, vector<T>(m, zero()));
for (int i = 0; i < n; i++)
for (int j = 0; j < m; j++)
for (int k = 0; k < p; k++)
C[i][j] = (C[i][j] + (*this)[i][k] * B[k][j]);
A.swap(C);
return (*this);
}
Matrix &operator^=(long long k) {
Matrix B = Matrix::I(height());
while (k > 0) {
if (k & 1)
B *= *this;
*this *= *this;
k >>= 1LL;
}
A.swap(B.A);
return (*this);
}
Matrix operator+(const Matrix &B) const { return (Matrix(*this) += B); }
Matrix operator-(const Matrix &B) const { return (Matrix(*this) -= B); }
Matrix operator*(const Matrix &B) const { return (Matrix(*this) *= B); }
Matrix operator^(const long long k) const { return (Matrix(*this) ^= k); }
bool operator==(const Matrix &B) const {
size_t n = height(), m = width();
if (n != B.height() || m != B.width())
return false;
for (int i = 0; i < n; i++)
for (int j = 0; j < m; j++)
if ((*this)[i][j] != B[i][j])
return false;
return true;
}
friend ostream &operator<<(ostream &os, Matrix &p) {
size_t n = p.height(), m = p.width();
for (int i = 0; i < n; i++) {
for (int j = 0; j < m; j++) {
os << p[i][j] << (j + 1 == m ? "\n" : " ");
}
}
return (os);
}
T determinant() { // O(n^3)
Matrix B(*this);
assert(width() == height());
T ret = 1;
for (int i = 0; i < width(); i++) {
int idx = -1;
for (int j = i; j < width(); j++) {
if (B[j][i] != 0)
idx = j;
}
if (idx == -1)
return (0);
if (i != idx) {
ret *= -1;
swap(B[i], B[idx]);
}
ret *= B[i][i];
T vv = B[i][i];
for (int j = 0; j < width(); j++) {
B[i][j] /= vv;
}
for (int j = i + 1; j < width(); j++) {
T a = B[j][i];
for (int k = 0; k < width(); k++) {
B[j][k] -= B[i][k] * a;
}
}
}
return (ret);
}
};
void solve(ll n, int k) {
Matrix<mint> C(k + 2);
rep(i, 0, k + 1) rep(j, 0, k + 1) C[i][j] = comb(k - i, k - j - i);
C[k + 1][k + 1] = 1;
C[k + 1][0] = 1;
C ^= n - 1;
mint ans = 0;
rep(i, 0, k + 2) ans += C[k + 1][i];
cout << ans.val() << endl;
}
int main() {
ll n;
int k;
cin >> n >> k;
solve(n, k);
}
SnowBeenDiding