結果

問題 No.2837 Flip Triomino
コンテスト
ユーザー 👑 binap
提出日時 2024-08-09 22:19:16
言語 C++17
(gcc 13.3.0 + boost 1.87.0)
結果
AC  
実行時間 107 ms / 2,000 ms
コード長 3,822 bytes
コンパイル時間 3,660 ms
コンパイル使用メモリ 258,228 KB
最終ジャッジ日時 2025-02-23 21:41:32
ジャッジサーバーID
（参考情報） 		judge5 / judge1
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 * 34
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 vector<int> vi;
typedef vector<long long> vl;
typedef vector<vector<int>> vvi;
typedef vector<vector<long long>> vvl;
typedef long double ld;
typedef pair<int, int> P;

ostream& operator<<(ostream& os, const modint& a) {os << a.val(); return os;}
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<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;

// combination mod prime
// https://youtu.be/8uowVvQ_-Mo?t=6002
// https://youtu.be/Tgd_zLfRZOQ?t=9928
struct modinv {
  int n; vector<mint> d;
  modinv(): n(2), d({0,1}) {}
  mint operator()(int i) {
    while (n <= i) d.push_back(-d[mint::mod()%n]*(mint::mod()/n)), ++n;
    return d[i];
  }
  mint operator[](int i) const { return d[i];}
} invs;
struct modfact {
  int n; vector<mint> d;
  modfact(): n(2), d({1,1}) {}
  mint operator()(int i) {
    while (n <= i) d.push_back(d.back()*n), ++n;
    return d[i];
  }
  mint operator[](int i) const { return d[i];}
} facts;
struct modfactinv {
  int n; vector<mint> d;
  modfactinv(): n(2), d({1,1}) {}
  mint operator()(int i) {
    while (n <= i) d.push_back(d.back()*invs(n)), ++n;
    return d[i];
  }
  mint operator[](int i) const { return d[i];}
} ifacts;
mint comb(int n, int k) {
  if (n < k || k < 0) return 0;
  return facts(n)*ifacts(k)*ifacts(n-k);
}

int main(){
	int n, m;
	cin >> n >> m;
	vector<string> a(n);
	cin >> a;
	vector<vector<int>> v(3, vector<int>(3));
	rep(i, n) rep(j, m){
		int idx = -1;
		if(a[i][j] == 'W') idx = 0;
		if(a[i][j] == 'B') idx = 1;
		if(a[i][j] == '?') idx = 2;
		v[(i + j) % 3][idx]++;
	}
	mint ans1 = 1;
	rep(i, 3){
		int x = v[i][2];
		mint tmp = 0;
		if(v[i][1] % 2){
			for(int y = 1; y <= x; y += 2) tmp += comb(x, y);
		}else{
			for(int y = 0; y <= x; y += 2) tmp += comb(x, y);
		}
		ans1 *= tmp;
	}
	mint ans2 = 1;
	rep(i, 3){
		int x = v[i][2];
		mint tmp = 0;
		if(v[i][1] % 2){
			for(int y = 0; y <= x; y += 2) tmp += comb(x, y);
		}else{
			for(int y = 1; y <= x; y += 2) tmp += comb(x, y);
		}
		ans2 *= tmp;
	}
	cout << ans1 + ans2 << "\n";
	return 0;
}
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