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
コンテスト
ユーザー maru143
提出日時 2020-08-26 14:27:15
言語 C++14
(gcc 13.3.0 + boost 1.87.0)
結果
WA  
実行時間 -
コード長 6,205 bytes
コンパイル時間 1,774 ms
コンパイル使用メモリ 174,540 KB
実行使用メモリ 5,248 KB
最終ジャッジ日時 2024-11-07 08:08:45
合計ジャッジ時間 2,579 ms
ジャッジサーバーID
（参考情報） 		judge3 / judge2
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 * 10 WA * 2
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 # 

#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using pii = pair<int, int>;
using pll = pair<ll, ll>;
#define rep(i, n) for(int i = 0; i < n; i++)
#define REP(i, a, b) for(int i = a; a < b; i++);
#define all(x) x.begin(), x.end()
#define rall(x) x.rbegin(), x.rend()
#define fi first
#define se second
#define pb push_back
#define debug(x) cerr << #x << ": " << x << endl
#define debug_vec(v) cerr << #v << ":"; rep(i, v.size()) cerr << " " << v[i]; cerr << endl
template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return 1; } return 0; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return 1; } return 0; }
const int INF = (1<<30) - 1;
const ll LINF = (1LL<<60) - 1;

template<class T>
struct Matrix {
    vector<vector<T>> A;

    Matrix() {}

    Matrix(size_t n, size_t m, T num) : A(n, vector<T>(m, num)) {}

    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);
    }

    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);
    }
};

ll mod;

struct ModInt {
    int x;

    ModInt() : x(0) {}

    ModInt(int64_t y) : x(y >= 0 ? y % mod : (mod - (-y) % mod) % mod) {}

    ModInt &operator+=(const ModInt &p) {
        if((x += p.x) >= mod) x -= mod;
        return *this;
    }

    ModInt &operator-=(const ModInt &p) {
        if((x += mod - p.x) >= mod) x -= mod;
        return *this;
    }

    ModInt &operator*=(const ModInt &p) {
        x = (int) (1LL * x * p.x % mod);
        return *this;
    }

    ModInt &operator/=(const ModInt &p) {
        *this *= p.inverse();
        return *this;
    }

    ModInt operator-() const { return ModInt(-x); }

    ModInt operator+(const ModInt &p) const { return ModInt(*this) += p; }

    ModInt operator-(const ModInt &p) const { return ModInt(*this) -= p; }

    ModInt operator*(const ModInt &p) const { return ModInt(*this) *= p; }

    ModInt operator/(const ModInt &p) const { return ModInt(*this) /= p; }

    bool operator==(const ModInt &p) const { return x == p.x; }

    bool operator!=(const ModInt &p) const { return x != p.x; }

    ModInt inverse() const {
        int a = x, b = mod, u = 1, v = 0, t;
        while(b > 0) {
            t = a / b;
            swap(a -= t * b, b);
            swap(u -= t * v, v);
        }
        return ModInt(u);
    }

    ModInt pow(int64_t n) const {
        ModInt ret(1), mul(x);
        while(n > 0) {
            if(n & 1) ret *= mul;
            mul *= mul;
            n >>= 1;
        }
        return ret;
    }

    friend ostream &operator<<(ostream &os, const ModInt &p) {
        return os << p.x;
    }

    friend istream &operator>>(istream &is, ModInt &a) {
        int64_t t;
        is >> t;
        a = ModInt(t);
        return (is);
    }

    static int get_mod() { return mod; }
};

using Mint = ModInt;

void solve() {
    ll n;
    cin >> n >> mod;
    Matrix<Mint> A(2);
    A[0][0] = 1; A[0][1] = 1;
    A[1][0] = 1; A[1][1] = 0;
    cout << (A ^ (n - 1))[1][0] << endl;
}

int main(){
    cin.tie(0);
    ios::sync_with_stdio(false);
    solve();
    return 0;
}
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