結果

問題 No.891 隣接3項間の漸化式
ユーザー a
提出日時 2019-09-20 22:06:48
言語 C++14
(gcc 13.3.0 + boost 1.87.0)
結果
AC  
実行時間 2 ms / 2,000 ms
コード長 5,708 bytes
コンパイル時間 1,976 ms
コンパイル使用メモリ 176,816 KB
実行使用メモリ 5,376 KB
最終ジャッジ日時 2024-09-14 17:44:25
合計ジャッジ時間 3,192 ms
ジャッジサーバーID
(参考情報)
judge4 / judge5
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 * 39
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"
using namespace std;
using namespace std;
template <class T>
struct Matrix
{
vector<vector<T>> A;
Matrix(size_t n) : A(n, vector<T>(n, 0)) {}
Matrix(size_t h, size_t w) : A(h, vector<T>(w, 0)) {}
Matrix(vector<vector<T>> &X) : A(X) {}
inline vector<T> &operator[](int idx) { return A[idx]; }
inline const vector<T> &operator[](int idx) const { return A[idx]; }
size_t height() const { return A.size(); }
size_t width() const { return A[0].size(); }
//ex)auto E = Matrix<int>::E(3);
static Matrix E(int n)
{
Matrix e(n);
for (int i = 0; i < n; i++)
e[i][i] = 1;
return e;
}
Matrix operator+(const Matrix &B) const
{
size_t h = this->height(), w = this->width();
assert(h == B.height() && w == B.width());
Matrix C(h, w);
for (int i = 0; i < h; i++)
for (int j = 0; j < w; j++)
{
C[i][j] = (*this)[i][j] + B[i][j];
}
return C;
}
Matrix operator-(const Matrix &B)
{
size_t h = this->height(), w = this->width();
assert(h == B.height() && w == B.width());
Matrix C(h, w);
for (int i = 0; i < h; i++)
for (int j = 0; j < w; j++)
{
C[i][j] = (*this)[i][j] - B[i][j];
}
return C;
}
Matrix operator*(const Matrix &B)
{
size_t h = this->height(), x = this->width(), w = B.width();
assert(x == B.height());
Matrix C(h, w);
for (size_t i = 0; i < h; i++)
for (size_t k = 0; k < x; k++)
for (size_t j = 0; j < w; j++)
{
C[i][j] += (*this)[i][k] * B[k][j];
}
return C;
}
Matrix &operator+=(const Matrix &B) { return (*this) = (*this) + B; }
Matrix &operator-=(const Matrix &B) { return (*this) = (*this) - B; }
Matrix &operator*=(const Matrix &B) { return (*this) = (*this) * B; }
Matrix power(long k)
{
auto n = this->height();
assert(k >= 0 && this->width() == n);
auto R = Matrix<T>::E(n), C = Matrix(*this);
while (k)
{
if (k & 1)
R *= C;
C *= C;
k >>= 1;
}
return R;
}
friend ostream &operator<<(ostream &o, const Matrix &A)
{
for (int i = 0; i < A.height(); i++)
{
for (int j = 0; j < A.width(); j++)
{
o << A[i][j] << " ";
}
o << endl;
}
return o;
}
};
template <int p>
struct Modint
{
int value;
Modint() : value(0) {}
Modint(long x) : value(x >= 0 ? x % p : (p + x % p) % p) {}
inline Modint &operator+=(const Modint &b)
{
if ((this->value += b.value) >= p)
this->value -= p;
return (*this);
}
inline Modint &operator-=(const Modint &b)
{
if ((this->value += p - b.value) >= p)
this->value -= p;
return (*this);
}
inline Modint &operator*=(const Modint &b)
{
this->value = (int)((1LL * this->value * b.value) % p);
return (*this);
}
inline Modint &operator/=(const Modint &b)
{
(*this) *= b.inverse();
return (*this);
}
Modint operator+(const Modint &b) const { return Modint(*this) += b; }
Modint operator-(const Modint &b) const { return Modint(*this) -= b; }
Modint operator*(const Modint &b) const { return Modint(*this) *= b; }
Modint operator/(const Modint &b) const { return Modint(*this) /= b; }
inline Modint &operator++(int) { return (*this) += 1; }
inline Modint &operator--(int) { return (*this) -= 1; }
inline bool operator==(const Modint &b) const
{
return this->value == b.value;
}
inline bool operator!=(const Modint &b) const
{
return this->value != b.value;
}
inline bool operator<(const Modint &b) const
{
return this->value < b.value;
}
inline bool operator<=(const Modint &b) const
{
return this->value <= b.value;
}
inline bool operator>(const Modint &b) const
{
return this->value > b.value;
}
inline bool operator>=(const Modint &b) const
{
return this->value >= b.value;
}
// requires that "this->value and p are co-prime"
// a_i * v + a_(i+1) * p = r_i
// r_i = r_(i+1) * q_(i+1) * r_(i+2)
// q == 1 (i > 1)
// reference: https://atcoder.jp/contests/agc026/submissions/2845729
// (line:93)
inline Modint inverse() const
{
assert(this->value != 0);
int r0 = p, r1 = this->value, a0 = 0, a1 = 1;
while (r1)
{
int q = r0 / r1;
r0 -= q * r1;
swap(r0, r1);
a0 -= q * a1;
swap(a0, a1);
}
return Modint(a0);
}
friend istream &operator>>(istream &is, Modint<p> &a)
{
long t;
is >> t;
a = Modint<p>(t);
return is;
}
friend ostream &operator<<(ostream &os, const Modint<p> &a)
{
return os << a.value;
}
};
/*
verified @ https://atcoder.jp/contests/abc034/submissions/4316971
*/
const int MOD = 1e9 + 7;
using Int = Modint<MOD>;
void solve()
{
Int a, b, n;
cin >> a >> b >> n;
if (n.value == 0)
{
cout << 0 << endl;
return;
}
Matrix<Int> v(2, 1), m(2, 2);
v[0][0] = 1;
m[0][0] = a;
m[0][1] = b;
m[1][0] = 1;
m = m.power(n.value - 1);
v = m * v;
cout << v[0][0] << endl;
}
int main()
{
solve();
return 0;
}
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