結果
問題 | No.526 フィボナッチ数列の第N項をMで割った余りを求める |
ユーザー | nanophoto12 |
提出日時 | 2020-08-22 06:22:03 |
言語 | C++17 (gcc 12.3.0 + boost 1.83.0) |
結果 |
AC
|
実行時間 | 2 ms / 2,000 ms |
コード長 | 4,258 bytes |
コンパイル時間 | 2,385 ms |
コンパイル使用メモリ | 210,648 KB |
実行使用メモリ | 6,820 KB |
最終ジャッジ日時 | 2024-10-15 06:45:58 |
合計ジャッジ時間 | 2,993 ms |
ジャッジサーバーID (参考情報) |
judge3 / judge2 |
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テストケース
テストケース表示入力 | 結果 | 実行時間 実行使用メモリ |
---|---|---|
testcase_00 | AC | 2 ms
6,816 KB |
testcase_01 | AC | 2 ms
6,820 KB |
testcase_02 | AC | 2 ms
6,820 KB |
testcase_03 | AC | 2 ms
6,816 KB |
testcase_04 | AC | 2 ms
6,816 KB |
testcase_05 | AC | 2 ms
6,816 KB |
testcase_06 | AC | 2 ms
6,816 KB |
testcase_07 | AC | 2 ms
6,816 KB |
testcase_08 | AC | 2 ms
6,820 KB |
testcase_09 | AC | 2 ms
6,816 KB |
testcase_10 | AC | 2 ms
6,820 KB |
testcase_11 | AC | 2 ms
6,816 KB |
testcase_12 | AC | 2 ms
6,816 KB |
testcase_13 | AC | 2 ms
6,820 KB |
testcase_14 | AC | 2 ms
6,820 KB |
ソースコード
#include <bits/stdc++.h> #define M_PI 3.14159265358979323846 // pi using namespace std; typedef long long ll; typedef pair<ll, ll> P; typedef tuple<ll, ll, ll> t3; #define rep(a,n) for(ll a = 0;a < n;a++) static const ll INF = 1e15; ll mod = 1e9+7; template<typename T> static inline void chmin(T& ref, const T value) { if (ref > value) ref = value; } template<typename T> static inline void chmax(T& ref, const T value) { if (ref < value) ref = value; } 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 Identity(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(); 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(); 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]); C[i][j] %= mod; } A.swap(C); return (*this); } Matrix& operator^=(long long k) { Matrix B = Matrix::Identity(height()); while (k > 0) { if (k & 1) B *= *this; *this *= *this; k >>= 1LL; } A.swap(B.A); return (*this); } Matrix& apply(std::function<T(T)> operation) { size_t n = height(), m = width(); for (int i = 0; i < n; i++) for (int j = 0; j < m; j++) (*this)[i][j] = operation((*this)[i][j]); return (*this); } Matrix& pow(long long k) { Matrix B = Matrix::Identity(height()); while (k > 0) { if (k & 1) { B *= *this; B.apply([&](long long v) {return v % mod; }); } *this *= *this; apply([&](long long v) {return v % mod; }); k >>= 1LL; } A.swap(B.A); return (*this); } vector<T> product(const vector<T>& right) { assert(width() == right.size()); vector<T> left(height()); for (int y = 0; y < height(); y++) { T sum = 0; for (int x = 0; x < width(); x++) { sum += this[y][x] * right[x]; } left[y] = sum; } return left; } 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); } friend ostream& operator<<(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 + 1 == m ? "]\n" : ","); } } return (os); } 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() { //F2 = F1 + F0 //F3 = F1 + F0 + F0; //F3 = 1 2 x F1 //F2 = 1 1 F0 //F3 = 1 1 x F2 //F2 = 1 0 F1 ll n, m; cin >> n >> m; Matrix<ll> mat(2, 2); mat[0][0] = 1; mat[0][1] = 1; mat[1][0] = 1; mod = m; auto p = mat.pow(n-1); cout << p[1][0] << endl; return 0; }