結果

問題 No.1521 Playing Musical Chairs Alone
ユーザー nok0
提出日時 2021-05-05 00:08:42
言語 C++17
(gcc 13.3.0 + boost 1.87.0)
結果
AC  
実行時間 108 ms / 2,000 ms
コード長 4,556 bytes
コンパイル時間 2,225 ms
コンパイル使用メモリ 203,528 KB
最終ジャッジ日時 2025-01-21 07:00:45
ジャッジサーバーID
(参考情報)
judge5 / judge3
このコードへのチャレンジ
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ファイルパターン 結果
sample AC * 3
other AC * 23
権限があれば一括ダウンロードができます

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/modint>
using namespace std;
#pragma region datastructure Matrix
#include <cassert>
#include <iostream>
#include <vector>
template <class T>
struct Matrix {
private:
std::vector<std::vector<T>> A;
static Matrix I(size_t n) {
Matrix mat(n);
for(int i = 0; i < n; i++) mat[i][i] = 1;
return mat;
}
public:
Matrix() = default;
Matrix(std::vector<std::vector<T>> &vvec) { A = vvec; }
Matrix(size_t n, size_t m) : A(n, std::vector<T>(m, 0)) {}
Matrix(size_t n, size_t m, T init) : A(n, std::vector<T>(m, init)) {}
Matrix(size_t n, std::vector<T> &vec) : A(n, vec) {}
Matrix(size_t n) : A(n, std::vector<T>(n, 0)) {}
size_t height() const { return A.size(); }
size_t width() const { return A[0].size(); }
inline const std::vector<T> &operator[](int k) const {
return A[k];
}
inline std::vector<T> &operator[](int k) {
return A[k];
}
Matrix &operator+=(const Matrix &B) {
size_t n = height(), m = width();
assert(n == B.height() and 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() and 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());
std::vector<std::vector<T>> C(n, std::vector<T>(m, 0));
for(int i = 0; i < n; i++)
for(int j = 0; j < m; j++)
for(int k = 0; k < p; k++)
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;
}
bool operator==(const Matrix &B) {
size_t n = height(), m = width();
if(n != B.height() or 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;
}
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);
}
Matrix &operator+=(const T &t) {
int n = height(), m = width();
for(int i = 0; i < n; i++)
for(int j = 0; j < m; j++)
(*this)[i][j] += t;
return *this;
}
Matrix &operator-=(const T &t) {
int n = height(), m = width();
for(int i = 0; i < n; i++)
for(int j = 0; j < m; j++)
(*this)[i][j] -= t;
return *this;
}
Matrix &operator*=(const T &t) {
int n = height(), m = width();
for(int i = 0; i < n; i++)
for(int j = 0; j < m; j++)
(*this)[i][j] *= t;
return *this;
}
Matrix &operator/=(const T &t) {
int n = height(), m = width();
for(int i = 0; i < n; i++)
for(int j = 0; j < m; j++)
(*this)[i][j] /= t;
return *this;
}
Matrix operator+(const T &t) const {
return (Matrix(*this) += t);
}
Matrix operator-(const T &t) const {
return (Matrix(*this) -= t);
}
Matrix operator*(const T &t) const {
return (Matrix(*this) *= t);
}
Matrix operator/(const T &t) const {
return (Matrix(*this) /= t);
}
friend std::ostream &operator<<(std::ostream &os, Matrix &p) {
size_t n = p.height(), m = p.width();
for(int i = 0; i < n; i++) {
os << '[';
for(int j = 0; j < m; j++)
os << p[i][j] << (j == m - 1 ? "]\n" : ",");
}
return (os);
}
T determinant() {
Matrix B(*this);
size_t n = height(), m = width();
assert(n == m);
T ret = 1;
for(int i = 0; i < n; i++) {
int idx = -1;
for(int j = i; j < n; 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 < n; j++) B[i][j] /= vv;
for(int j = i + 1; j < n; j++) {
T a = B[j][i];
for(int k = 0; k < n; k++) {
B[j][k] -= B[i][k] * a;
}
}
}
return ret;
}
};
#pragma endregion
using mint = atcoder::modint1000000007;
int n, k, l;
int main() {
cin >> n >> k >> l;
Matrix<mint> mat(1, n), mul(n, n);
mat[0][0] = 1;
for(int i = 0; i < n; i++)
for(int j = 1; j <= l; j++)
mul[i][(i + j) % n] = 1;
mat *= mul ^ k;
for(int i = 0; i < n; i++)
cout << mat[0][i].val() << '\n';
return 0;
}
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