結果
問題 | No.2381 Gift Exchange Party |
ユーザー |
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提出日時 | 2023-07-14 21:52:02 |
言語 | C++11(廃止可能性あり) (gcc 13.3.0) |
結果 |
AC
|
実行時間 | 62 ms / 2,000 ms |
コード長 | 2,630 bytes |
コンパイル時間 | 1,652 ms |
コンパイル使用メモリ | 161,176 KB |
実行使用メモリ | 7,408 KB |
最終ジャッジ日時 | 2024-09-16 06:47:54 |
合計ジャッジ時間 | 4,065 ms |
ジャッジサーバーID (参考情報) |
judge3 / judge5 |
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ファイルパターン | 結果 |
---|---|
sample | AC * 3 |
other | AC * 22 |
ソースコード
#include <bits/stdc++.h>using namespace std;#define F0(i,n) for (int i=0; i<n; i++)#define F1(i,n) for (int i=1; i<=n; i++)#define CL(a,x) memset(x, a, sizeof(x));#define SZ(x) ((int)x.size())const int inf = 1000009;const double pi = acos(-1.0);typedef pair<int, int> pii;typedef long long ll;const double EPS = 1e-9;const int MOD = 998244353;#define PR(x) cerr << #x << "=" << x << endltemplate<class A, class B>ostream& operator<<(ostream& os, const pair<A, B>& p) { os << "(" << p.first << ", " << p.second << ")"; return os; }template<class A, class B, class C>ostream& operator<<(ostream& os, const tuple<A, B, C>& p) { os << "(" << get<0>(p) << ", " << get<1>(p) << ", " << get<2>(p) << ")"; return os; }istream& operator>>(istream& is, pii& p) { is>>p.first>>p.second; return is; }template<class T>ostream& operator<<(ostream& os, const vector<T>& v) { os << "["; F0(i,SZ(v)) { if (i>0) os << ","; os << v[i]; } os << "]"; return os; }template<class T>ostream& operator<<(ostream& os, const set<T>& v) { os << "{"; int f=1; for(auto i:v) { if(f)f=0;else os << ","; cerr << i; } os << "}" << endl;return os; }template<class T, class R>ostream& operator<<(ostream& os, const map<T,R>& v) { os << "{"; int f=1; for(auto i:v) { if(f)f=0;else os << ", "; cerr << i.first << ":" << i.second; } os << "}" << endl; return os; }int i, j, k, n;ll m, ans;const int N = 200005;const int M = 435;int dp[N];const int DX[]={-1,0,1,0};const int DY[]={0,1,0,-1};const string CS="nesw";const string HS="URDL";void Add(int& x, int y) {x += y;if (x >= MOD) x -= MOD;}ll Mult(int x, int y) {return 1LL * x * y % MOD;}ll modpow(ll x, ll n) {if (n == 0) return 1;ll y = modpow(x, n / 2); y = (y * y) % MOD;if (n & 1) y = (y * x) % MOD;return y;}const int MAXN = 200001;ll f[MAXN], rf[MAXN];ll C(int n, int k) {return f[n] * rf[k] % MOD * rf[n - k] % MOD;}void Solve() {dp[0] = 1;F1(i, n) {dp[i] = dp[i - 1];if (i >= m) {ll x = C(i - 1, m - 1);x = Mult(x, f[m - 1]);x = Mult(x, dp[i - m]);Add(dp[i], x);}}ans = f[n] - dp[n];if (ans < 0) ans += MOD;cout << ans << endl;}int main() {//ignore = freopen("x.in", "r", stdin);//freopen("x.out", "w", stdout);//int tn; cin >> tn;//F1(ti, tn) {f[0] = rf[0] = 1;for (int i = 1; i <= 200000; i++) {f[i] = (f[i - 1] * i) % MOD;rf[i] = modpow(f[i], MOD - 2);}while (cin >> n >> m) {//PR(n);Solve();}return 0;}