結果
問題 | No.2668 Trees on Graph Paper |
ユーザー | Misuki |
提出日時 | 2024-03-08 23:46:58 |
言語 | C++23 (gcc 13.3.0 + boost 1.87.0) |
結果 |
AC
|
実行時間 | 1,970 ms / 3,000 ms |
コード長 | 4,564 bytes |
コンパイル時間 | 2,245 ms |
コンパイル使用メモリ | 199,656 KB |
実行使用メモリ | 237,692 KB |
最終ジャッジ日時 | 2024-09-29 20:54:41 |
合計ジャッジ時間 | 24,305 ms |
ジャッジサーバーID (参考情報) |
judge1 / judge4 |
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ファイルパターン | 結果 |
---|---|
other | AC * 33 |
ソースコード
#pragma GCC optimize("O2")#include <algorithm>#include <array>#include <bit>#include <bitset>#include <cassert>#include <cctype>#include <cfenv>#include <cfloat>#include <chrono>#include <cinttypes>#include <climits>#include <cmath>#include <compare>#include <complex>#include <concepts>#include <cstdarg>#include <cstddef>#include <cstdint>#include <cstdio>#include <cstdlib>#include <cstring>#include <deque>#include <fstream>#include <functional>#include <initializer_list>#include <iomanip>#include <ios>#include <iostream>#include <istream>#include <iterator>#include <limits>#include <list>#include <map>#include <memory>#include <new>#include <numbers>#include <numeric>#include <ostream>#include <queue>#include <random>#include <ranges>#include <set>#include <span>#include <sstream>#include <stack>#include <streambuf>#include <string>#include <tuple>#include <type_traits>#include <variant>#define int ll#define INT128_MAX (__int128)(((unsigned __int128) 1 << ((sizeof(__int128) * __CHAR_BIT__) - 1)) - 1)#define INT128_MIN (-INT128_MAX - 1)#define clock chrono::steady_clock::now().time_since_epoch().count()#ifdef DEBUG#define dbg(x) cout << (#x) << " = " << x << '\n'#else#define dbg(x)#endifnamespace R = std::ranges;namespace V = std::views;using namespace std;using ll = long long;using ull = unsigned long long;using ldb = long double;using pii = pair<int, int>;using pll = pair<ll, ll>;//#define double ldbtemplate<class T>ostream& operator<<(ostream& os, const pair<T, T> pr) {return os << pr.first << ' ' << pr.second;}template<class T, size_t N>ostream& operator<<(ostream& os, const array<T, N> &arr) {for(const T &X : arr)os << X << ' ';return os;}template<class T>ostream& operator<<(ostream& os, const vector<T> &vec) {for(const T &X : vec)os << X << ' ';return os;}template<class T>ostream& operator<<(ostream& os, const set<T> &s) {for(const T &x : s)os << x << ' ';return os;}//note: inversion only works when MOD is a primestruct mint {static long long MOD;long long _val;mint(long long init = 0) {_val = init % MOD;(*this).norm();}mint POW(long long index) {if (index == 0)return mint(1ll);mint base = *this;mint res = (base == 0ll ? 0ll : 1ll);while(index) {if (index & 1)res *= base;base *= base, index >>= 1;}return res;}mint inv() { return (*this).POW(MOD - 2); }mint& norm() {if (_val >= MOD)_val -= MOD;if (_val < 0)_val += MOD;return *this;}mint& operator+=(mint b) {_val += b._val;return (*this).norm();}mint& operator-=(mint b) {_val -= b._val;return (*this).norm();}mint& operator*=(mint b) {_val = (_val * b._val) % MOD;return *this;}mint& operator/=(mint b) {_val = (_val * b.inv()._val) % MOD;return *this;}mint& operator++() {_val += 1;return (*this).norm();}mint& operator--() {_val -= 1;return (*this).norm();}mint operator++(signed) {mint tmp = *this;++(*this);return tmp;}mint operator--(signed) {mint tmp = *this;--(*this);return tmp;}mint operator-() { return mint(-_val); }bool operator==(mint b) { return _val == b._val; }bool operator!=(mint b) { return _val != b._val; }friend mint operator+(mint a, mint b) { return a += b; }friend mint operator-(mint a, mint b) { return a -= b; }friend mint operator*(mint a, mint b) { return a *= b; }friend mint operator/(mint a, mint b) { return a /= b; }friend ostream& operator<<(ostream& os, const mint& b) {return os << b._val;}friend istream& operator>>(istream& is, mint& b) {long long val;is >> val;b = mint(val);return is;}};ll mint::MOD = 2;signed main() {ios::sync_with_stdio(false), cin.tie(NULL);int n; cin >> n >> mint::MOD;vector<array<mint, 3>> dp(2 * n);dp[1] = {1, 1, 1};for(int i = 1; i < 2 * n - 3; i++) {for(int j : {0, 1, 2}) {dp[i + 1][j] += dp[i][0];dp[i + 2][j] += dp[i][1] * (i + 1);dp[i + 3][j] += dp[i][2] * (i + 1) * (i + 2);}for(int j : {1, 2}) {dp[i + 1][j] += dp[i][1];dp[i + 2][j] += dp[i][2] * (i + 1);}dp[i + 1][2] += dp[i][2];}mint ans = 1;for(int i = 2; i <= (2 * n - 1); i++) ans *= i;for(int i = 1; i <= 2 * n - 3; i += 2) ans *= dp[i][2];cout << ans << '\n';return 0;}