結果
問題 | No.891 隣接3項間の漸化式 |
ユーザー |
|
提出日時 | 2022-11-14 16:16:06 |
言語 | C++17 (gcc 13.3.0 + boost 1.87.0) |
結果 |
AC
|
実行時間 | 2 ms / 2,000 ms |
コード長 | 3,087 bytes |
コンパイル時間 | 4,608 ms |
コンパイル使用メモリ | 295,112 KB |
最終ジャッジ日時 | 2025-02-08 20:13:37 |
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 * 39 |
ソースコード
#pragma GCC optimize("O3")#pragma GCC optimize("unroll-loops")#include<bits/stdc++.h>using namespace std;typedef long long ll;#include<atcoder/all>using namespace atcoder;using mint=modint1000000007;template< class T >struct Matrix {vector< vector< T > > A;Matrix() {}Matrix(size_t n, size_t m) : A(n, vector< T >(m, 0)) {}Matrix(size_t n) : A(n, vector< T >(n, 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, 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] = (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);}T determinant() {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);}};int main(){cin.tie(0);ios::sync_with_stdio(false);ll a,b,n;cin >> a >> b >> n;Matrix<mint> x(2,2),y(2,1);y[0][0]=0;y[1][0]=1;x[0][1]=1;x[1][1]=a;x[1][0]=b;auto res=(x^n)*y;cout << res[0][0].val() << "\n";}