結果

問題 No.1302 Random Tree Score
ユーザー yuto1115
提出日時 2020-11-27 22:51:51
言語 C++17
(gcc 13.3.0 + boost 1.87.0)
結果
AC  
実行時間 2,148 ms / 3,000 ms
コード長 7,093 bytes
コンパイル時間 2,690 ms
コンパイル使用メモリ 207,000 KB
最終ジャッジ日時 2025-01-16 08:13:31
ジャッジサーバーID
(参考情報)
judge5 / judge2
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ファイルパターン 結果
sample AC * 3
other AC * 14
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ソースコード

diff #
プレゼンテーションモードにする

//@formatter:off
#include<bits/stdc++.h>
#define rep(i,n) for (int i = 0; i < int(n); ++i)
#define rrep(i,n) for (int i = int(n)-1; i >= 0; i--)
#define rep2(i,s,n) for (int i = int(s); i < int(n); ++i)
#define all(a) a.begin(),a.end()
#define rall(a) a.rbegin(),a.rend()
#define pb push_back
#define eb emplace_back
#define vi vector<int>
#define vvi vector<vector<int>>
#define vl vector<ll>
#define vvl vector<vector<ll>>
#define vd vector<double>
#define vvd vector<vector<double>>
#define vs vector<string>
#define vc vector<char>
#define vvc vector<vector<char>>
#define vb vector<bool>
#define vvb vector<vector<bool>>
#define vp vector<P>
#define vvp vector<vector<P>>
using namespace std;
using ll = long long;
using P = pair<int,int>;
using LP = pair<ll,ll>;
template<class S,class T> istream& operator>>(istream &is,pair<S,T> &p) { return is >> p.first >> p.second; }
template<class S,class T> ostream& operator<<(ostream &os,const pair<S,T> &p) { return os<<'{'<<p.first<<","<<p.second<<'}'; }
template<class T> istream& operator>>(istream &is,vector<T> &v) { for(T &t:v){is>>t;} return is; }
template<class T> ostream& operator<<(ostream &os,const vector<T> &v) { os<<'[';rep(i,v.size())os<<v[i]<<(i==int(v.size()-1)?"":","); return os<<']';
    }
void Yes(bool b) { cout << (b ? "Yes" : "No") << '\n'; }
void YES(bool b) { cout << (b ? "YES" : "NO") << '\n'; }
template<class T> void fin(T a) { cout << a << endl; exit(0); }
template<class T> bool chmin(T& a,T b) {if(a > b){a = b; return true;} return false;}
template<class T> bool chmax(T& a,T b) {if(a < b){a = b; return true;} return false;}
const int inf = 1001001001;
const ll linf = 1001001001001001001;
//@formatter:on
//constexpr int mod = 1000000007;
constexpr int mod = 998244353;
struct mint {
ll x;
constexpr mint(ll x = 0) : x((x % mod + mod) % mod) {}
constexpr mint operator-() const { return mint(-x); }
constexpr mint &operator+=(const mint &a) {
if ((x += a.x) >= mod) x -= mod;
return *this;
}
constexpr mint &operator++() { return *this += mint(1); }
constexpr mint &operator-=(const mint &a) {
if ((x += mod - a.x) >= mod) x -= mod;
return *this;
}
constexpr mint &operator--() { return *this -= mint(1); }
constexpr mint &operator*=(const mint &a) {
(x *= a.x) %= mod;
return *this;
}
constexpr mint operator+(const mint &a) const {
mint res(*this);
return res += a;
}
constexpr mint operator-(const mint &a) const {
mint res(*this);
return res -= a;
}
constexpr mint operator*(const mint &a) const {
mint res(*this);
return res *= a;
}
constexpr mint pow(ll t) const {
mint res = mint(1), a(*this);
while (t > 0) {
if (t & 1) res *= a;
t >>= 1;
a *= a;
}
return res;
}
// for prime mod
constexpr mint inv() const { return pow(mod - 2); }
constexpr mint &operator/=(const mint &a) { return *this *= a.inv(); }
constexpr mint operator/(const mint &a) const {
mint res(*this);
return res /= a;
}
};
ostream &operator<<(ostream &os, const mint &a) { return os << a.x; }
bool operator==(const mint &a, const mint &b) { return a.x == b.x; }
bool operator!=(const mint &a, const mint &b) { return a.x != b.x; }
bool operator==(const mint &a, const int &b) { return a.x == b; }
bool operator!=(const mint &a, const int &b) { return a.x != b; }
template< typename Mint >
struct NumberTheoreticTransformFriendlyModInt {
vector< int > rev;
vector< Mint > rts;
int base, max_base;
Mint root;
NumberTheoreticTransformFriendlyModInt() : base(1), rev{0, 1}, rts{0, 1} {
assert(mod >= 3 && mod % 2 == 1);
auto tmp = mod - 1;
max_base = 0;
while(tmp % 2 == 0) tmp >>= 1, max_base++;
root = 2;
while(root.pow((mod - 1) >> 1) == 1) root += 1;
assert(root.pow(mod - 1) == 1);
root = root.pow((mod - 1) >> max_base);
}
void ensure_base(int nbase) {
if(nbase <= base) return;
rev.resize(1 << nbase);
rts.resize(1 << nbase);
for(int i = 0; i < (1 << nbase); i++) {
rev[i] = (rev[i >> 1] >> 1) + ((i & 1) << (nbase - 1));
}
assert(nbase <= max_base);
while(base < nbase) {
Mint z = root.pow(1 << (max_base - 1 - base));
for(int i = 1 << (base - 1); i < (1 << base); i++) {
rts[i << 1] = rts[i];
rts[(i << 1) + 1] = rts[i] * z;
}
++base;
}
}
void ntt(vector< Mint > &a) {
const int n = (int) a.size();
assert((n & (n - 1)) == 0);
int zeros = __builtin_ctz(n);
ensure_base(zeros);
int shift = base - zeros;
for(int i = 0; i < n; i++) {
if(i < (rev[i] >> shift)) {
swap(a[i], a[rev[i] >> shift]);
}
}
for(int k = 1; k < n; k <<= 1) {
for(int i = 0; i < n; i += 2 * k) {
for(int j = 0; j < k; j++) {
Mint z = a[i + j + k] * rts[j + k];
a[i + j + k] = a[i + j] - z;
a[i + j] = a[i + j] + z;
}
}
}
}
void intt(vector< Mint > &a) {
const int n = (int) a.size();
ntt(a);
reverse(a.begin() + 1, a.end());
Mint inv_sz = Mint(1) / n;
for(int i = 0; i < n; i++) a[i] *= inv_sz;
}
vector< Mint > multiply(vector< Mint > a, vector< Mint > b) {
int need = a.size() + b.size() - 1;
int nbase = 1;
while((1 << nbase) < need) nbase++;
ensure_base(nbase);
int sz = 1 << nbase;
a.resize(sz, 0);
b.resize(sz, 0);
ntt(a);
ntt(b);
Mint inv_sz = Mint(1) / sz;
for(int i = 0; i < sz; i++) {
a[i] *= b[i] * inv_sz;
}
reverse(a.begin() + 1, a.end());
ntt(a);
a.resize(need);
return a;
}
};
NumberTheoreticTransformFriendlyModInt<mint> ntt;
vector<mint> pow(const vector<mint> &f, int k) {
int n = f.size();
vector<mint> res(n, 0);
res[0] = 1;
vector<mint> now = f;
while (k > 0) {
if (k & 1) {
res = ntt.multiply(res, now);
if (res.size() > n) res.erase(res.begin() + n, res.end());
}
now = ntt.multiply(now, now);
if (now.size() > n) now.erase(now.begin() + n, now.end());
k >>= 1;
}
return res;
}
int main() {
ios::sync_with_stdio(false);
cin.tie(nullptr);
cout.tie(nullptr);
int n;
cin >> n;
vector<mint> v(2 * n - 1, 0);
mint now = 1;
rep2(i, 1, n) {
if (i > 1) now /= i - 1;
v[i] = now * i;
}
auto ans = pow(v, n);
now = 1;
rep2(i, 1, n - 1) now *= i;
cout << ans[2 * n - 2] * now / mint(n).pow(n - 2) << endl;
}
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