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 | KY2001 |
提出日時 | 2021-05-05 22:55:32 |
言語 | C++17 (gcc 12.3.0 + boost 1.83.0) |
結果 |
WA
|
実行時間 | - |
コード長 | 5,187 bytes |
コンパイル時間 | 2,428 ms |
コンパイル使用メモリ | 215,812 KB |
実行使用メモリ | 6,948 KB |
最終ジャッジ日時 | 2024-09-13 15:21:12 |
合計ジャッジ時間 | 3,190 ms |
ジャッジサーバーID (参考情報) |
judge2 / judge4 |
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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 | WA | - |
testcase_01 | WA | - |
testcase_02 | WA | - |
testcase_03 | WA | - |
testcase_04 | WA | - |
testcase_05 | WA | - |
testcase_06 | WA | - |
testcase_07 | WA | - |
testcase_08 | WA | - |
testcase_09 | WA | - |
testcase_10 | WA | - |
testcase_11 | WA | - |
testcase_12 | WA | - |
testcase_13 | WA | - |
testcase_14 | WA | - |
ソースコード
//@Author: KeinYukiyoshi // clang-format off #include <bits/stdc++.h> //#pragma GCC optimize("Ofast") //#pragma GCC target("avx") #define int long long using namespace std; #define stoi stoll #define fi first #define se second #define rep(i, n) for(int i=0, i##_len=(n); i<i##_len; i++) #define rep2(i, a, b) for (int i = (int)(a), i##_len=(b); i < i##_len; i++) #define rep3(i, a, b) for (int i = (int)(a), i##_len=(b); i >= i##_len; i--) #define FOR(i, a) for (auto &i: a) #define ALL(obj) begin(obj), end(obj) #define _max(x) *max_element(ALL(x)) #define _min(x) *min_element(ALL(x)) #define _sum(x) accumulate(ALL(x), 0LL) int MOD = 1000000007; // const int MOD = 998244353; // const int INF = (int)1e18; // const int INF = 10000000000007; // 1e13 + 7 // const int INF = LLONG_MAX; // 9.2e18 const double EPS = 1e-8; const double PI = 3.14159265358979; template <class T> using V = vector<T>; template <class T> using VV = vector<vector<T>>; template <class T> using VVV = vector<vector<vector<T>>>; template <class T, class S> using P = pair<T, S>; template<class T> bool chmax(T &a, const T &b) {if (a < b) {a = b;return true;}return false;} template<class T> bool chmin(T &a, const T &b) {if (b < a) {a = b;return true;}return false;} int _ceil(int a, int b) { return (a >= 0 ? (a + (b - 1)) / b : (a - (b - 1)) / b); } template<class T> T chmod(T &a, T mod=MOD) {a = a >= 0 ? a % mod : a - (mod * _ceil(a, mod)); return a;}; int _mod(int a, int mod=MOD) {return a >= 0 ? a % mod : a - (mod * _ceil(a, mod));} int _pow(int a, int b) {int res = 1;for (a %= MOD; b; a = a * a % MOD, b >>= 1)if (b & 1) res = res * a % MOD;return res;} struct mint {long long x;mint(long long x = 0) : x((x % MOD + MOD) % MOD) {}mint operator-() const { return mint(-x); }mint &operator+=(const mint a) {if ((x += a.x) >= MOD) x -= MOD;return *this;}mint &operator-=(const mint a) {if ((x += MOD - a.x) >= MOD) x -= MOD;return *this;}mint &operator*=(const mint a) { (x *= a.x) %= MOD;return *this; }mint operator+(const mint a) const { return mint(*this) += a; }mint operator-(const mint a) const { return mint(*this) -= a; } mint operator*(const mint a) const { return mint(*this) *= a; }mint pow(long long t) const {if (!t) return 1;mint a = pow(t >> 1);a *= a;if (t & 1) a *= *this;return a;}mint inv() const { return pow(MOD - 2); }mint &operator/=(const mint a) { return *this *= a.inv(); }mint operator/(const mint a) const { return mint(*this) /= a; }};istream &operator>>(istream &is, mint &a) { return is >> a.x; }ostream &operator<<(ostream &os, const mint &a) { return os << a.x; } // clang-format on template <class T> struct matrix { int row; // 行 int column; // 列 vector<vector<T>> M; matrix(vector<vector<T>> &m) : row(m.size()), column(m[0].size()) { M = m; } matrix add(matrix &m) { // M += m for (int i = 0; i < row; i++) for (int j = 0; j < column; j++) M[i][j] += m.M[i][j]; return M; } matrix subtract(matrix &m) { // M -= m for (int i = 0; i < row; i++) for (int j = 0; j < column; j++) M[i][j] -= m.M[i][j]; return M; } void multiply_from_right(matrix &m) { //M = M*m O(n^3) vector<vector<T>> temp(row, vector<T>(m.column)); for (int i = 0; i < row; i++) for (int j = 0; j < m.column; j++) for (int k = 0; k < column; k++) { temp[i][j] += M[i][k] * m.M[k][j]; } M = temp, row = M.size(), column = M[0].size(); } void multiply_from_left(matrix &m) { //M = m*M O(n^3) vector<vector<T>> temp(m.row, vector<T>(column)); for (int i = 0; i < m.row; i++) for (int j = 0; j < column; j++) for (int k = 0; k < m.column; k++) temp[i][j] += m.M[i][k] * M[k][j]; M = temp, row = M.size(), column = M[0].size(); } void pow(int b) { // M = pow(M, n) O(n^3log b) int n = row; vector<vector<T>> ret(n, vector<T>(n)); for (int i = 0; i < n; i++) ret[i][i] = 1; for (; b; b >>= 1) { if (b & 1) { vector<vector<T>> temp(n, vector<T>(n)); for (int i = 0; i < n; i++) for (int j = 0; j < n; j++) for (int k = 0; k < n; k++) temp[i][j] += ret[i][k] * M[k][j]; ret = temp; } vector<vector<T>> temp(row, vector<T>(column)); for (int i = 0; i < row; i++) for (int j = 0; j < column; j++) for (int k = 0; k < column; k++) temp[i][j] += M[i][k] * M[k][j]; M = temp; } M = ret; } void show() { for (int i = 0; i < row; i++) { for (int j = 0; j < column; j++) cout << M[i][j] << " "; cout << endl; } } }; class No526NM { public: static void solve(istream &cin, ostream &cout) { cin.tie(nullptr); cout.tie(nullptr); ios::sync_with_stdio(false); cout << fixed << setprecision(15); int N, M; cin >> N >> M; MOD = M; VV<mint> A = {{0, 1}, {1, 1}}; VV<mint> temp = {{0}, {1}}; matrix<mint> a(A); matrix<mint> b(temp); a.pow(N - 2); a.show(); b.multiply_from_left(a); cout << b.M[1][0] << endl; } }; signed main() { No526NM solver; std::istream& in(std::cin); std::ostream& out(std::cout); solver.solve(in, out); return 0; }