結果

問題 No.2964 Obstruction Bingo
コンテスト
ユーザー 👑 binap
提出日時 2024-11-16 17:32:49
言語 C++17
(gcc 13.3.0 + boost 1.87.0)
結果
WA  
実行時間 -
コード長 3,363 bytes
コンパイル時間 4,396 ms
コンパイル使用メモリ 260,524 KB
最終ジャッジ日時 2025-02-25 04:47:11
ジャッジサーバーID
（参考情報） 		judge4 / judge2
このコードへのチャレンジ
（要ログイン）
ファイルパターン 結果
sample AC * 3
other AC * 5 WA * 44
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;

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;

int main(){
	int l, k;
	cin >> l >> k;
	string s, t;
	cin >> s >> t;
	int n = (k + 1) * (k + 1);
	vector<int> a(26);
	cin >> a;
	int tot = 0;
	rep(i, 26) tot += a[i];
	vector<map<int, mint>> G(n);
	auto f = [&](int x, int y){
		return x * (k + 1) + y;
	};
	for(int x_from = 0; x_from < l; x_from++){
		for(int y_from = 0; y_from < l; y_from++){
			int from = f(x_from, y_from);
			rep(i, 26){
				int x_to = x_from;
				int y_to = y_from;
				if(s[x_from % l] == 'a' + i) x_to++;
				if(t[y_from % l] == 'a' + i) y_to++;
				int to = f(x_to, y_to);
				G[from][to] += mint(a[i]) / tot;
			}
		}
	}
	vector<mint> dp(n);
	dp[0] = 1;
	mint ans1 = 0;
	mint ans2 = 0;
	rep(_, k){
		vector<mint> dp_old(n);
		swap(dp, dp_old);
		rep(from, n){
			for(auto [to, c] : G[from]) dp[to] += dp_old[from] * c;
		}
		rep(to, n){
			int x_to = to / (k + 1);
			int y_to = to % (k + 1);
			if(x_to == y_to + l){
				ans1 += dp[to];
				dp[to] = 0;
			}
			if(x_to + l == y_to){
				ans2 += dp[to];
				dp[to] = 0;
			}
		}
	}
	cout << ans1 << ' ' << ans2 << "\n";
	return 0;
}
0