結果
問題 | No.2503 Typical Path Counting Problem on a Grid |
ユーザー |
![]() |
提出日時 | 2023-10-12 01:11:02 |
言語 | C++17 (gcc 13.3.0 + boost 1.87.0) |
結果 |
AC
|
実行時間 | 477 ms / 2,000 ms |
コード長 | 10,973 bytes |
コンパイル時間 | 2,676 ms |
コンパイル使用メモリ | 244,012 KB |
最終ジャッジ日時 | 2025-02-17 06:46:12 |
ジャッジサーバーID (参考情報) |
judge2 / judge2 |
(要ログイン)
ファイルパターン | 結果 |
---|---|
other | AC * 10 |
ソースコード
// #define _GLIBCXX_DEBUG#pragma GCC optimize("O2,no-stack-protector,unroll-loops,fast-math")#include <bits/stdc++.h>using namespace std;#define rep(i, n) for (int i = 0; i < int(n); i++)#define per(i, n) for (int i = (n)-1; 0 <= i; i--)#define rep2(i, l, r) for (int i = (l); i < int(r); i++)#define per2(i, l, r) for (int i = (r)-1; int(l) <= i; i--)#define each(e, v) for (auto& e : v)#define MM << " " <<#define pb push_back#define eb emplace_back#define all(x) begin(x), end(x)#define rall(x) rbegin(x), rend(x)#define sz(x) (int)x.size()template <typename T> void print(const vector<T>& v, T x = 0) {int n = v.size();for (int i = 0; i < n; i++) cout << v[i] + x << (i == n - 1 ? '\n' : ' ');if (v.empty()) cout << '\n';}using ll = long long;using pii = pair<int, int>;using pll = pair<ll, ll>;template <typename T> bool chmax(T& x, const T& y) {return (x < y) ? (x = y, true) : false;}template <typename T> bool chmin(T& x, const T& y) {return (x > y) ? (x = y, true) : false;}template <class T>using minheap = std::priority_queue<T, std::vector<T>, std::greater<T>>;template <class T> using maxheap = std::priority_queue<T>;template <typename T> int lb(const vector<T>& v, T x) {return lower_bound(begin(v), end(v), x) - begin(v);}template <typename T> int ub(const vector<T>& v, T x) {return upper_bound(begin(v), end(v), x) - begin(v);}template <typename T> void rearrange(vector<T>& v) {sort(begin(v), end(v));v.erase(unique(begin(v), end(v)), end(v));}// __int128_t gcd(__int128_t a, __int128_t b) {// if (a == 0)// return b;// if (b == 0)// return a;// __int128_t cnt = a % b;// while (cnt != 0) {// a = b;// b = cnt;// cnt = a % b;// }// return b;// }struct Union_Find_Tree {vector<int> data;const int n;int cnt;Union_Find_Tree(int n) : data(n, -1), n(n), cnt(n) {}int root(int x) {if (data[x] < 0) return x;return data[x] = root(data[x]);}int operator[](int i) { return root(i); }bool unite(int x, int y) {x = root(x), y = root(y);if (x == y) return false;if (data[x] > data[y]) swap(x, y);data[x] += data[y], data[y] = x;cnt--;return true;}int size(int x) { return -data[root(x)]; }int count() { return cnt; };bool same(int x, int y) { return root(x) == root(y); }void clear() {cnt = n;fill(begin(data), end(data), -1);}};template <int mod> struct Mod_Int {int x;Mod_Int() : x(0) {}Mod_Int(long long y) : x(y >= 0 ? y % mod : (mod - (-y) % mod) % mod) {}static int get_mod() { return mod; }Mod_Int& operator+=(const Mod_Int& p) {if ((x += p.x) >= mod) x -= mod;return *this;}Mod_Int& operator-=(const Mod_Int& p) {if ((x += mod - p.x) >= mod) x -= mod;return *this;}Mod_Int& operator*=(const Mod_Int& p) {x = (int)(1LL * x * p.x % mod);return *this;}Mod_Int& operator/=(const Mod_Int& p) {*this *= p.inverse();return *this;}Mod_Int& operator++() { return *this += Mod_Int(1); }Mod_Int operator++(int) {Mod_Int tmp = *this;++*this;return tmp;}Mod_Int& operator--() { return *this -= Mod_Int(1); }Mod_Int operator--(int) {Mod_Int tmp = *this;--*this;return tmp;}Mod_Int operator-() const { return Mod_Int(-x); }Mod_Int operator+(const Mod_Int& p) const { return Mod_Int(*this) += p; }Mod_Int operator-(const Mod_Int& p) const { return Mod_Int(*this) -= p; }Mod_Int operator*(const Mod_Int& p) const { return Mod_Int(*this) *= p; }Mod_Int operator/(const Mod_Int& p) const { return Mod_Int(*this) /= p; }bool operator==(const Mod_Int& p) const { return x == p.x; }bool operator!=(const Mod_Int& p) const { return x != p.x; }Mod_Int inverse() const {assert(*this != Mod_Int(0));return pow(mod - 2);}Mod_Int pow(long long k) const {Mod_Int now = *this, ret = 1;for (; k > 0; k >>= 1, now *= now) {if (k & 1) ret *= now;}return ret;}friend ostream& operator<<(ostream& os, const Mod_Int& p) {return os << p.x;}friend istream& operator>>(istream& is, Mod_Int& p) {long long a;is >> a;p = Mod_Int<mod>(a);return is;}};ll mpow2(ll x, ll n, ll mod) {ll ans = 1;x %= mod;while (n != 0) {if (n & 1) ans = ans * x % mod;x = x * x % mod;n = n >> 1;}ans %= mod;return ans;}template <typename T> T modinv(T a, const T& m) {T b = m, u = 1, v = 0;while (b > 0) {T t = a / b;swap(a -= t * b, b);swap(u -= t * v, v);}return u >= 0 ? u % m : (m - (-u) % m) % m;}ll divide_int(ll a, ll b) {if (b < 0) a = -a, b = -b;return (a >= 0 ? a / b : (a - b + 1) / b);}// const int MOD = 1000000007;const int MOD = 998244353;using mint = Mod_Int<MOD>;// ----- library -------template <typename T>struct Matrix {vector<vector<T>> A;int n, m;Matrix(int n, int m) : A(n, vector<T>(m, 0)), n(n), m(m) {}inline const vector<T> &operator[](int k) const { return A[k]; }inline vector<T> &operator[](int k) { return A[k]; }static Matrix I(int l) {Matrix ret(l, l);for (int i = 0; i < l; i++) ret[i][i] = 1;return ret;}Matrix &operator*=(const Matrix &B) {assert(m == B.n);Matrix ret(n, B.m);for (int i = 0; i < n; i++) {for (int k = 0; k < m; k++) {for (int j = 0; j < B.m; j++) ret[i][j] += A[i][k] * B[k][j];}}swap(A, ret.A);m = B.m;return *this;}Matrix operator*(const Matrix &B) const { return Matrix(*this) *= B; }Matrix pow(long long k) const {assert(n == m);Matrix now = *this, ret = I(n);for (; k > 0; k >>= 1, now *= now) {if (k & 1) ret *= now;}return ret;}bool eq(const T &a, const T &b) const {return a == b;// return abs(a-b) <= EPS;}// 行基本変形を用いて簡約化を行い、(rank, det) の組を返すpair<int, T> row_reduction(vector<T> &b) {assert((int)b.size() == n);if (n == 0) return make_pair(0, m > 0 ? 0 : 1);int check = 0, rank = 0;T det = (n == m ? 1 : 0);for (int j = 0; j < m; j++) {int pivot = check;for (int i = check; i < n; i++) {if (A[i][j] != 0) pivot = i;// if(abs(A[i][j]) > abs(A[pivot][j])) pivot = i; // T が小数の場合はこちら}if (check != pivot) det *= T(-1);swap(A[check], A[pivot]), swap(b[check], b[pivot]);if (eq(A[check][j], T(0))) {det = T(0);continue;}rank++;det *= A[check][j];T r = T(1) / A[check][j];for (int k = j + 1; k < m; k++) A[check][k] *= r;b[check] *= r;A[check][j] = T(1);for (int i = 0; i < n; i++) {if (i == check) continue;if (!eq(A[i][j], 0)) {for (int k = j + 1; k < m; k++) A[i][k] -= A[i][j] * A[check][k];b[i] -= A[i][j] * b[check];}A[i][j] = T(0);}if (++check == n) break;}return make_pair(rank, det);}pair<int, T> row_reduction() {vector<T> b(n, T(0));return row_reduction(b);}// 行基本変形を行い、逆行列を求めるpair<bool, Matrix> inverse() {if (n != m) return make_pair(false, Matrix(0, 0));if (n == 0) return make_pair(true, Matrix(0, 0));Matrix ret = I(n);for (int j = 0; j < n; j++) {int pivot = j;for (int i = j; i < n; i++) {if (A[i][j] != 0) pivot = i;// if(abs(A[i][j]) > abs(A[pivot][j])) pivot = i; // T が小数の場合はこちら}swap(A[j], A[pivot]), swap(ret[j], ret[pivot]);if (eq(A[j][j], T(0))) return make_pair(false, Matrix(0, 0));T r = T(1) / A[j][j];for (int k = j + 1; k < n; k++) A[j][k] *= r;for (int k = 0; k < n; k++) ret[j][k] *= r;A[j][j] = T(1);for (int i = 0; i < n; i++) {if (i == j) continue;if (!eq(A[i][j], T(0))) {for (int k = j + 1; k < n; k++) A[i][k] -= A[i][j] * A[j][k];for (int k = 0; k < n; k++) ret[i][k] -= A[i][j] * ret[j][k];}A[i][j] = T(0);}}return make_pair(true, ret);}// Ax = b の解の 1 つと解空間の基底の組を返すvector<vector<T>> Gaussian_elimination(vector<T> b) {row_reduction(b);vector<vector<T>> ret;vector<int> p(n, m);vector<bool> is_zero(m, true);for (int i = 0; i < n; i++) {for (int j = 0; j < m; j++) {if (!eq(A[i][j], T(0))) {p[i] = j;break;}}if (p[i] < m) {is_zero[p[i]] = false;} else if (!eq(b[i], T(0))) {return {};}}vector<T> x(m, T(0));for (int i = 0; i < n; i++) {if (p[i] < m) x[p[i]] = b[i];}ret.push_back(x);for (int j = 0; j < m; j++) {if (!is_zero[j]) continue;x[j] = T(1);for (int i = 0; i < n; i++) {if (p[i] < m) x[p[i]] = -A[i][j];}ret.push_back(x);x[j] = T(0);}return ret;}};// ----- library -------int main() {ios::sync_with_stdio(false);std::cin.tie(nullptr);cout << fixed << setprecision(15);const int si = 1e7 + 10;vector<mint> f(si);f[0] = 1, f[1] = 2;rep2(i, 2, si) f[i] = f[i - 1] * i * 2 + f[i - 2] * (i - 1);int T;cin >> T;while (T--) {ll n, m;cin >> n >> m;if (n > m)swap(n, m);if (n == 0) {cout << 1 << '\n';continue;}Matrix<mint> a(2, 2), b(2, 1);a[0][0] = n * 2 + 1, a[0][1] = n;a[1][0] = 1, a[1][1] = 0;b[0][0] = f[n];b[1][0] = f[n - 1];auto ret = a.pow(m - n) * b;cout << ret[0][0] * f[n] + ret[1][0] * f[n - 1] * n << '\n';}}