結果

問題 No.2983 Christmas Color Grid (Advent Calender ver.)
ユーザー KumaTachiRenKumaTachiRen
提出日時 2024-12-08 02:19:55
言語 C++23
(gcc 13.3.0 + boost 1.87.0)
結果
TLE  
実行時間 -
コード長 5,604 bytes
コンパイル時間 3,119 ms
コンパイル使用メモリ 261,848 KB
実行使用メモリ 14,592 KB
最終ジャッジ日時 2024-12-08 02:20:47
合計ジャッジ時間 46,600 ms
ジャッジサーバーID
(参考情報)
judge4 / judge5
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 * 60 TLE * 4
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 #
プレゼンテーションモードにする

#include <bits/stdc++.h>
using namespace std;
#include <atcoder/modint>
using namespace atcoder;
using mint = modint998244353;
using vm = vector<mint>;
using vvm = vector<vm>;
inline ostream &operator<<(ostream &os, const mint x)
{
return os << x.val();
};
inline istream &operator>>(istream &is, mint &x)
{
long long v;
is >> v;
x = v;
return is;
};
struct Fast
{
Fast()
{
std::cin.tie(nullptr);
ios::sync_with_stdio(false);
cout << setprecision(10);
}
} fast;
#define all(a) (a).begin(), (a).end()
#define contains(a, x) ((a).find(x) != (a).end())
#define rep(i, a, b) for (int i = (a); i < (int)(b); i++)
#define rrep(i, a, b) for (int i = (int)(b) - 1; i >= (a); i--)
#define YN(b) cout << ((b) ? "YES" : "NO") << "\n";
#define Yn(b) cout << ((b) ? "Yes" : "No") << "\n";
#define yn(b) cout << ((b) ? "yes" : "no") << "\n";
template <class T>
inline bool chmin(T &a, T b)
{
if (a > b)
{
a = b;
return true;
}
return false;
}
template <class T>
inline bool chmax(T &a, T b)
{
if (a < b)
{
a = b;
return true;
}
return false;
}
using ll = long long;
using vb = vector<bool>;
using vvb = vector<vb>;
using vi = vector<int>;
using vvi = vector<vi>;
using vl = vector<ll>;
using vvl = vector<vl>;
template <typename T1, typename T2>
ostream &operator<<(ostream &os, pair<T1, T2> &p)
{
os << "(" << p.first << "," << p.second << ")";
return os;
}
template <typename T>
ostream &operator<<(ostream &os, vector<T> &vec)
{
for (int i = 0; i < (int)vec.size(); i++)
{
os << vec[i] << (i + 1 == (int)vec.size() ? "" : " ");
}
return os;
}
template <typename T>
istream &operator>>(istream &is, vector<T> &vec)
{
for (int i = 0; i < (int)vec.size(); i++)
is >> vec[i];
return is;
}
ll floor(ll a, ll b) { return a >= 0 ? a / b : (a + 1) / b - 1; }
ll ceil(ll a, ll b) { return a > 0 ? (a - 1) / b + 1 : a / b; }
int dx[] = {1, 0, -1, 0};
int dy[] = {0, 1, 0, -1};
void solve()
{
int h, w, m;
ll k;
cin >> h >> w >> k >> m;
if (h < w) swap(h, w);
using mint = dynamic_modint<0>;
mint::set_mod(m);
vector<int> rev(1 << w, 0);
rep(v, 0, 1 << w) rep(i, 0, w) rev[v] |= ((v >> i) & 1) << (w - 1 - i);
int mask = (1 << w) - 1;
vector<mint> cnt(h * w + 1, 0);
vector<bool> ok(1 << (h * w), false);
int vc[] = {0, 0, 0, 0};
rep(bit, 1, 1 << (h * w))
{
int rbit = 0;
rep(i, 0, h) rbit |= rev[(bit >> (i * w)) & mask] << (i * w);
int cbit = 0;
rep(i, 0, h) cbit |= ((bit >> (i * w)) & mask) << ((h - 1 - i) * w);
int rcbit = 0;
rep(i, 0, h) rcbit |= rev[(bit >> (i * w)) & mask] << ((h - 1 - i) * w);
if (bit <= rbit && bit <= cbit && bit <= rcbit)
{
int v = 0, cm = 0;
rep(i, 0, h + 1) rep(j, 0, w)
{
int u = w * i + j;
int z = (v >> (j * 2)) & 3;
if ((bit >> u) & 1)
{
if (z)
{
int y = j ? ((v >> (j * 2 - 2)) & 3) : 0;
if (y > 0 && y != z)
{
rep(x, 0, w) if (((v >> (x * 2)) & 3) == y)
{
v ^= ((y ^ z) << (x * 2));
vc[y]--;
vc[z]++;
}
}
}
else
{
int y = j ? (v >> (j * 2 - 2)) & 3 : 0;
if (y)
{
z = y;
}
else
{
z = 1;
while (vc[z])
z++;
}
vc[z]++;
v |= z << (j * 2);
}
}
else
{
if (vc[z] == 1) cm++;
vc[z]--;
v &= ~(3 << (j * 2));
}
}
ok[bit] = cm == 1;
ok[rbit] = ok[bit];
ok[cbit] = ok[bit];
ok[rcbit] = ok[bit];
if (!ok[bit]) continue;
set<int> st;
st.insert(bit);
st.insert(rbit);
st.insert(cbit);
st.insert(rcbit);
int val = (int)st.size();
int pc = __builtin_popcount(bit);
int c = 0;
for (int x = 0; x < h; x++)
for (int y = 0; y < w; y++)
{
if ((bit >> (w * x + y)) & 1) continue;
bool f = false;
for (int i = 0; i < 4; i++)
{
int nx = x + dx[i], ny = y + dy[i];
if (nx < 0 || nx >= h || ny < 0 || ny >= w) continue;
f |= (bit >> (w * nx + ny)) & 1;
}
if (f) c++;
}
cnt[pc] += val << (h * w - pc - c);
}
}
mint ans = 0;
rep(i, 0, h * w + 1) ans += mint(i).pow(k) * cnt[i];
mint p = 0;
rep(i, 1, h * w + 1) p += mint(i).inv();
ans *= p;
ans *= mint(2).inv().pow(h * w);
cout << ans.val() << "\n";
}
int main()
{
int t = 1;
// cin >> t;
while (t--)
solve();
}
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