結果

問題 No.2837 Flip Triomino
ユーザー dadasdadas
提出日時 2024-08-09 22:47:54
言語 C++17
(gcc 12.3.0 + boost 1.83.0)
結果
WA  
実行時間 -
コード長 3,494 bytes
コンパイル時間 2,222 ms
コンパイル使用メモリ 206,148 KB
実行使用メモリ 7,648 KB
最終ジャッジ日時 2024-08-09 22:47:58
合計ジャッジ時間 4,021 ms
ジャッジサーバーID
（参考情報） 		judge3 / judge1
このコードへのチャレンジ
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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 AC 12 ms
7,260 KB
testcase_01 WA -
testcase_02 WA -
testcase_03 AC 12 ms
7,392 KB
testcase_04 WA -
testcase_05 WA -
testcase_06 WA -
testcase_07 AC 13 ms
7,520 KB
testcase_08 WA -
testcase_09 WA -
testcase_10 WA -
testcase_11 WA -
testcase_12 WA -
testcase_13 WA -
testcase_14 WA -
testcase_15 WA -
testcase_16 WA -
testcase_17 AC 13 ms
7,516 KB
testcase_18 AC 12 ms
7,516 KB
testcase_19 WA -
testcase_20 WA -
testcase_21 WA -
testcase_22 WA -
testcase_23 WA -
testcase_24 WA -
testcase_25 WA -
testcase_26 WA -
testcase_27 WA -
testcase_28 WA -
testcase_29 WA -
testcase_30 WA -
testcase_31 AC 13 ms
7,388 KB
testcase_32 WA -
testcase_33 WA -
testcase_34 WA -
testcase_35 WA -
testcase_36 AC 12 ms
7,388 KB
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;
void _main();
int main() {
	cin.tie(0);
	ios::sync_with_stdio(false);
	_main();
	return 0;
}
using ll = long long;
using ull = unsigned long long;
using ld = long double;
using vi = std::vector<int>;
using vvi = std::vector<vi>;
using vl = std::vector<ll>;
using vii = std::vector<pair<int, int> >;
using vvl = std::vector<vl>;
using vll = std::vector<pair<ll , ll> >;
using vd = std::vector<double>;
using vvd = std::vector<vd>;
using vs = std::vector<std::string>;
using vvs = std::vector<vs>;
using vb = std::vector<bool>;
using vvb = std::vector<vb>;
using vc = std::vector<char>;
using vvc = std::vector<vc>;
using pii = std::pair<int, int>;
using pll = std::pair<ll, ll>;
using piil = std::pair<pair<int, int>, ll>;
using mii = std::map<int, int>;
using mll = std::map<ll, ll>;
using pql = std::priority_queue<ll>;
using pqi = std::priority_queue<int>;
using pqiil = std::priority_queue<pair<pair<int, int>, ll> >;
using pqii = std::priority_queue<pair<int, int> >;

#define pb push_back
#define ps push
#define eb emplace_back
#define is insert
#define er erase
#define f first
#define s second
#define lb lower_bound
#define ub upper_bound
#define sf(i) sizeof(i)
#define endl "\n"
#define sz(v) ((int)(v).size())
#define all(v) (v).begin(), (v).end()
#define rep(i, L, R) for(ll i = L;i<=R;i++)
#define pcis precision

template<typename T>
struct infinity {
	static constexpr T max=std::numeric_limits<T>::max();
	static constexpr T min=std::numeric_limits<T>::min();
	static constexpr T value=std::numeric_limits<T>::max()/2;
	static constexpr T mvalue=std::numeric_limits<T>::min()/2;
};
template<typename T>constexpr T INF=infinity<T>::value;
constexpr ll lINF=INF<ll>;
constexpr int iINF = INF<int>;
constexpr ld PI = 3.1415926535897932384626;

ll MOD = 998244353;

ll frac[250001], inv[250001];
ll MD = MOD;

ll fpow(ll x, ll y, ll mod) {
	if (!y) return 1%mod;
	x %= mod;
	ll ret = 1;
	while(y > 0) {
		if(y%2 == 1) ret = (ret*x)%mod;
		x = (x*x)%mod;
		y /= 2;
	}
	return ret;
}

void initComb() {
	frac[0] = 1;
	inv[0] = 1;
	for (int i = 1; i <= 250000; i++) {
		frac[i] = (frac[i - 1] * i) % MD;
	}

	inv[250000] = fpow(frac[250000], MD - 2, MD);

	for (int i = 250000; i > 0; i--) {
		inv[i - 1] = (inv[i] * i) % MD;
	}
}

ll C(ll n, ll k) {
	if (n < k)return 0;
	return ((frac[n] * inv[k]) % MD * inv[n - k]) % MD;
}

void _main() {
	initComb();
	// 불변량?
	ll N, M;
	cin >>N >>M;
	string S[N+1];
	for (ll i = 1; i<=N ; i++)cin>>S[i];
	for (ll i = 1; i<=N ; i++)S[i] = " " + S[i];
	ll cnt0 = 0;
	ll cnt1 = 0;
	ll cnt2 = 0;

	ll SUM0 = 0;
	ll SUM1 = 0;
	ll SUM2 = 0;
	for (ll i = 1; i<=N; i++) {
		for (ll j= 1; j<=M; j++) {
			if (S[i][j] == '?') {
				if ((i+j)%3 == 0) {
					cnt0++;
				} else if ((i+j)%3 == 1) {
					cnt1++;
				} else if ((i+j)%3 == 2) {
					cnt2++;
				}
			} else if ('B') {
				if ((i+j)%3 == 0) {
					SUM0++;
				} else if ((i+j)%3 == 1) {
					SUM1++;
				} else if ((i+j)%3 == 2) {
					SUM2++;
				}
			}
		}
	}
	// SUM을 0으로 만들어야 함
	// 짝-홀을 SUM으로 만들어야 함
//	ll RES = fpow(2,cnt1+cnt2+cnt0-2, MOD);
	ll RES = 0;
	RES += (C(cnt0, (SUM0%2==0?cnt0/2+1:(cnt0+1)/2))*C(cnt1, (SUM1%2==0?cnt1/2+1:(cnt1+1)/2))%MOD)*C(cnt2,  (SUM2%2==0?cnt2/2+1:(cnt2+1)/2))%MOD;// 0 0 0
	RES += (C(cnt0, (SUM0%2!=0?cnt0/2+1:(cnt0+1)/2))*C(cnt1, (SUM1%2!=0?cnt1/2+1:(cnt1+1)/2))%MOD)*C(cnt2,  (SUM2%2!=0?cnt2/2+1:(cnt2+1)/2))%MOD;// 0 0 0
	RES %= MOD;
	cout << RES << endl;








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