結果
| 問題 |
No.1516 simple 門松列 problem Re:MASTER
|
| コンテスト | |
| ユーザー |
 fuppy_kyopro
|
| 提出日時 | 2021-05-21 23:27:34 |
| 言語 | C++17 (gcc 13.3.0 + boost 1.87.0) |
| 結果 |
TLE
(最新)
AC
(最初)
|
| 実行時間 | - |
| コード長 | 6,117 bytes |
| コンパイル時間 | 2,863 ms |
| コンパイル使用メモリ | 225,508 KB |
| 最終ジャッジ日時 | 2025-01-21 16:46:06 |
|
ジャッジサーバーID （参考情報） |
judge2 / judge5 |
（要ログイン）
| ファイルパターン | 結果 |
|---|---|
| sample | AC * 3 |
| other | AC * 18 TLE * 1 |
ソースコード
/*
#pragma GCC target("avx2")
#pragma GCC optimize("O3")
#pragma GCC optimize("unroll-loops")
//*/
#include <bits/stdc++.h>
// #include <atcoder/all>
using namespace std;
// using namespace atcoder;
#define DEBUG(x) cerr << #x << ": " << x << endl;
#define DEBUG_VEC(v) \
cerr << #v << ":"; \
for (int iiiiiiii = 0; iiiiiiii < v.size(); iiiiiiii++) \
cerr << " " << v[iiiiiiii]; \
cerr << endl;
#define DEBUG_MAT(v) \
cerr << #v << endl; \
for (int i = 0; i < v.size(); i++) { \
for (int j = 0; j < v[i].size(); j++) { \
cerr << v[i][j] << " "; \
} \
cerr << endl; \
}
typedef long long ll;
// #define int ll
#define vi vector<int>
#define vl vector<ll>
#define vii vector<vector<int>>
#define vll vector<vector<ll>>
#define vs vector<string>
#define pii pair<int, int>
#define pis pair<int, string>
#define psi pair<string, int>
#define pll pair<ll, ll>
template <class S, class T>
pair<S, T> operator+(const pair<S, T> &s, const pair<S, T> &t) {
return pair<S, T>(s.first + t.first, s.second + t.second);
}
template <class S, class T>
pair<S, T> operator-(const pair<S, T> &s, const pair<S, T> &t) { return pair<S, T>(s.first - t.first, s.second - t.second); }
template <class S, class T>
ostream &operator<<(ostream &os, pair<S, T> p) {
os << "(" << p.first << ", " << p.second << ")";
return os;
}
#define X first
#define Y second
#define rep(i, n) for (int i = 0; i < (int)(n); i++)
#define rep1(i, n) for (int i = 1; i <= (int)(n); i++)
#define rrep(i, n) for (int i = (int)(n)-1; i >= 0; i--)
#define rrep1(i, n) for (int i = (int)(n); i > 0; i--)
#define REP(i, a, b) for (int i = a; i < b; i++)
#define in(x, a, b) (a <= x && x < b)
#define all(c) c.begin(), c.end()
void YES(bool t = true) {
cout << (t ? "YES" : "NO") << endl;
}
void Yes(bool t = true) { cout << (t ? "Yes" : "No") << endl; }
void yes(bool t = true) { cout << (t ? "yes" : "no") << endl; }
void NO(bool t = true) { cout << (t ? "NO" : "YES") << endl; }
void No(bool t = true) { cout << (t ? "No" : "Yes") << endl; }
void no(bool t = true) { cout << (t ? "no" : "yes") << endl; }
template <class T>
bool chmax(T &a, const T &b) {
if (a < b) {
a = b;
return 1;
}
return 0;
}
template <class T>
bool chmin(T &a, const T &b) {
if (a > b) {
a = b;
return 1;
}
return 0;
}
#define UNIQUE(v) v.erase(std::unique(v.begin(), v.end()), v.end());
const ll inf = 1000000001;
const ll INF = (ll)1e18 + 1;
const long double pi = 3.1415926535897932384626433832795028841971L;
int popcount(ll t) { return __builtin_popcountll(t); }
// int dx[4] = {1, 0, -1, 0}, dy[4] = {0, 1, 0, -1};
// int dx2[8] = { 1,1,0,-1,-1,-1,0,1 }, dy2[8] = { 0,1,1,1,0,-1,-1,-1 };
vi dx = {0, 1, 0, -1}, dy = {-1, 0, 1, 0};
// vi dx2 = { 1,1,0,-1,-1,-1,0,1 }, dy2 = { 0,1,1,1,0,-1,-1,-1 };
struct Setup_io {
Setup_io() {
ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
cout << fixed << setprecision(25);
}
} setup_io;
// const ll MOD = 1000000007;
const ll MOD = 998244353;
// #define mp make_pair
//#define endl '\n'
using vlll = vector<vll>;
using vllll = vector<vlll>;
using P = pair<vllll, vllll>;
unordered_map<int, P> memo;
ll k;
P solve(int n) {
if (memo.count(n)) return memo[n];
DEBUG(n);
int l = n / 2;
int r = n - l;
auto dp1 = solve(l);
auto dp2 = solve(r);
memo[n].first = vllll(k, vlll(k, vll(k, vl(k))));
memo[n].second = vllll(k, vlll(k, vll(k, vl(k))));
rep(u1, k) rep(v1, k) rep(w1, k) rep(x1, k) {
if (dp1.first[u1][v1][w1][x1] == 0 && dp1.second[u1][v1][w1][x1] == 0) continue;
rep(u2, k) rep(v2, k) rep(w2, k) rep(x2, k) {
if (dp2.first[u2][v2][w2][x2] == 0 && dp2.second[u2][v2][w2][x2] == 0) continue;
if (x1 == u2) continue;
if (w1 == u2) continue;
if (x1 == v2) continue;
{
int mi = min({w1, x1, u2}), ma = max({w1, x1, u2});
if (mi != x1 && ma != x1) continue;
}
{
int mi = min({x1, u2, v2}), ma = max({x1, u2, v2});
if (mi != u2 && ma != u2) continue;
}
(memo[n].first[u1][v1][w2][x2] += dp1.first[u1][v1][w1][x1] * dp2.first[u2][v2][w2][x2]) %= MOD;
(memo[n].second[u1][v1][w2][x2] += dp1.first[u1][v1][w1][x1] * dp2.second[u2][v2][w2][x2] + dp1.second[u1][v1][w1][x1] * dp2.first[u2][v2][w2][x2]) %= MOD;
}
}
return memo[n];
}
void dfs(vi &state, int n) {
if (state.size() >= 3) {
int u = state[0], v = state[1], w = state[state.size() - 2], x = state[state.size() - 1];
int sum = 0;
rep(i, state.size()) sum += state[i];
memo[state.size()].first[u][v][w][x]++;
(memo[state.size()].second[u][v][w][x] += sum) %= MOD;
}
if (state.size() == min(n, 8)) {
return;
}
rep(x, k) {
if (state.size() > 0 && state.back() == x) continue;
if (state.size() > 1 && state[state.size() - 2] == x) continue;
if (state.size() > 1) {
int u = state[state.size() - 2], v = state[state.size() - 1];
int mi = min({u, v, x}), ma = max({u, v, x});
if (mi != v && ma != v) continue;
}
state.push_back(x);
dfs(state, n);
state.pop_back();
}
}
signed main() {
int n;
cin >> n >> k;
for (int x = 3; x <= 8; x++) {
memo[x].first = vllll(k, vlll(k, vll(k, vl(k))));
memo[x].second = vllll(k, vlll(k, vll(k, vl(k))));
}
vi state;
dfs(state, n);
P ans = solve(n);
ll ans1 = 0, ans2 = 0;
rep(i, k) rep(j, k) rep(l, k) rep(m, k) {
ans1 += ans.first[i][j][l][m];
ans2 += ans.second[i][j][l][m];
}
ans1 %= MOD;
ans2 %= MOD;
cout << ans1 << " " << ans2 << endl;
}