結果

問題 No.3255 01 Matrix Counting
ユーザー 👑 binap
提出日時 2025-09-05 23:03:47
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 3,821 bytes
コンパイル時間 4,820 ms
コンパイル使用メモリ 251,324 KB
実行使用メモリ 7,716 KB
最終ジャッジ日時 2025-09-05 23:04:06
合計ジャッジ時間 5,629 ms
ジャッジサーバーID
(参考情報)
judge3 / judge4
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 * 2
other AC * 14
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>
#include<atcoder/all>
#define rep(i,n) for(int i=0;i<n;i++)
using namespace std;
using namespace atcoder;
typedef long long ll;
typedef pair<int, int> P;

template <int m> ostream& operator<<(ostream& os, const static_modint<m>& a) {os << a.val(); return os;}
template <int m> ostream& operator<<(ostream& os, const dynamic_modint<m>& a) {os << a.val(); return os;}
template <int m> istream& operator>>(istream& is, static_modint<m>& a) {long long x; is >> x; a = x; return is;}
template <int m> istream& operator>>(istream& is, dynamic_modint<m>& a) {long long x; is >> x; a = x; return is;}
template<typename T> istream& operator>>(istream& is, vector<T>& v){int n = v.size(); assert(n > 0); rep(i, n) is >> v[i]; return is;}
template<typename U, typename T> ostream& operator<<(ostream& os, const pair<U, T>& p){os << p.first << ' ' << p.second; return os;}
template<typename T> ostream& operator<<(ostream& os, const vector<T>& v){int n = v.size(); rep(i, n) os << v[i] << (i == n - 1 ? "\n" : " "); return os;}
template<typename T> ostream& operator<<(ostream& os, const vector<vector<T>>& v){int n = v.size(); rep(i, n) os << v[i] << (i == n - 1 ? "\n" : ""); return os;}
template<typename T> ostream& operator<<(ostream& os, const set<T>& se){for(T x : se) os << x << " "; os << "\n"; return os;}
template<typename T> ostream& operator<<(ostream& os, const unordered_set<T>& se){for(T x : se) os << x << " "; os << "\n"; return os;}
template<typename S, auto op, auto e> ostream& operator<<(ostream& os, const atcoder::segtree<S, op, e>& seg){int n = seg.max_right(0, [](S){return true;}); rep(i, n) os << seg.get(i) << (i == n - 1 ? "\n" : " "); return os;}
template<typename S, auto op, auto e, typename F, auto mapping, auto composition, auto id> ostream& operator<<(ostream& os, const atcoder::lazy_segtree<S, op, e, F, mapping, composition, id>& seg){int n = seg.max_right(0, [](S){return true;}); rep(i, n) os << seg.get(i) << (i == n - 1 ? "\n" : " "); return os;}

template<typename T> void chmin(T& a, T b){a = min(a, b);}
template<typename T> void chmax(T& a, T b){a = max(a, b);}

using mint = modint998244353;

// https://youtu.be/ylWYSurx10A?t=2352
template<typename T>
struct Matrix  : vector<vector<T>> {
	int h, w;
	Matrix(int h, int w, T val=0): vector<vector<T>>(h, vector<T>(w, val)), h(h), w(w) {}
	Matrix(initializer_list<initializer_list<T>> a) : vector<vector<T>>(a.begin(), a.end()){
		assert(int(this->size()) >= 1);
		assert(int((*this)[0].size()) >= 1);
		h = this->size();
		w = (*this)[0].size();
		rep(i, h) assert(int((*this)[i].size()) == w);
	}
	Matrix& unit() {
		assert(h == w);
		rep(i,h) (*this)[i][i] = 1;
		return *this;
	}
	Matrix operator*=(const Matrix& M){
		assert(w == M.h);
		Matrix r(h, M.w);
		rep(i,h) rep(k,w) rep(j, M.w){
			r[i][j] += (*this)[i][k] * M[k][j];
		}
		swap(*this, r);
		return *this;
	}
	Matrix operator*(const Matrix& M) const {return (Matrix(*this) *= M);}
	Matrix operator*=(const T& a) {
		for(int i = 0; i < h; i++) for(int j = 0; j < w; j++) (*this)[i][j] *= a;
		return *this;
	}
	Matrix operator*(const T& a) const {return (Matrix(*this) *= a);}
	Matrix operator+=(const Matrix& M){
		assert(h == M.h and w == M.w);
		for(int i = 0; i < h; i++) for(int j = 0; j < w; j++) (*this)[i][j] += M[i][j];
		return *this;
	}
	Matrix operator+(const Matrix& M) const {return (Matrix(*this) += M);}
	Matrix pow(long long t) const {
		assert(h == w);
		if (!t) return Matrix(h,h).unit();
		if (t == 1) return *this;
		Matrix r = pow(t>>1);r = r*r;
		if (t&1) r = r*(*this);
		return r;
	}
};

int main(){
	int h, w;
	cin >> h >> w;
	auto e = [&](int n){
		if(n == 0) return mint(1);
		else return mint(2).pow(n - 1);
	};
	mint ans = 1;
	ans *= e(h).pow(w - 2);
	ans *= mint(2).pow(h - 2);
	cout << ans * 8 << "\n";
	return 0;
}
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