結果

問題 No.2068 Restricted Permutation
ユーザー PCTprobabilityPCTprobability
提出日時 2022-09-02 21:21:49
言語 C++17(gcc12)
(gcc 12.3.0 + boost 1.87.0)
結果
AC  
実行時間 139 ms / 2,000 ms
コード長 6,271 bytes
コンパイル時間 4,717 ms
コンパイル使用メモリ 274,368 KB
実行使用メモリ 81,164 KB
最終ジャッジ日時 2024-11-16 02:10:18
合計ジャッジ時間 8,142 ms
ジャッジサーバーID
（参考情報） 		judge2 / judge5
このコードへのチャレンジ
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ファイルパターン 結果
sample AC * 3
other AC * 23
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 # 

#include <bits/stdc++.h>
using namespace std;
#if __has_include(<atcoder/all>)
#include <atcoder/all>
using namespace atcoder;
#endif
using ll = long long;
using ld = long double;
using ull = unsigned long long;
#define endl "\n"
typedef pair<int, int> Pii;
#define REP(i, n) for (int i = 0; i < (n); ++i)
#define REP3(i, m, n) for (int i = (m); (i) < int(n); ++ (i))
#define FOR(i,a,b) for(ll i=a;i<=(ll)(b);i++)
#define rep(i,a,b) for(int i=(int)(a);i<(int)(b);i++)
#define ALL(x) begin(x), end(x)
#define rrep(i,a,b) for(int i=a;i>=b;i--)
#define fore(i,a) for(auto &i:a)
#define all(s) (s).begin(),(s).end()
#define drep2(i, m, n) for (int i = (m)-1; i >= (n); --i)
#define drep(i, n) drep2(i, n, 0)
#define rever(vec) reverse(vec.begin(), vec.end())
#define sor(vec) sort(vec.begin(), vec.end())
#define fi first
#define se second
#define pb push_back
#define P pair<ll,ll>
#define PQminll priority_queue<ll, vector<ll>, greater<ll>>
#define PQmaxll priority_queue<ll,vector<ll>,less<ll>>
#define PQminP priority_queue<P, vector<P>, greater<P>>
#define PQmaxP priority_queue<P,vector<P>,less<P>>
#define NP next_permutation
//const ll mod = 1000000009;
const ll mod = 998244353;
//const ll mod = 1000000007;
const ll inf = 4100000000000000000ll;
const ld eps = ld(0.00000000001);
//static const long double pi = 3.141592653589793;
template<class T>void vcin(vector<T> &n){for(int i=0;i<int(n.size());i++) cin>>n[i];}
template<class T,class K>void vcin(vector<T> &n,vector<K> &m){for(int i=0;i<int(n.size());i++) cin>>n[i]>>m[i];}
template<class T>void vcout(vector<T> &n){for(int i=0;i<int(n.size());i++){cout<<n[i]<<" ";}cout<<endl;}
template<class T>void vcin(vector<vector<T>> &n){for(int i=0;i<int(n.size());i++){for(int j=0;j<int(n[i].size());j++){cin>>n[i][j];}}}
template<class T>void vcout(vector<vector<T>> &n){for(int i=0;i<int(n.size());i++){for(int j=0;j<int(n[i].size());j++){cout<<n[i][j]<<" ";}cout<<endl;}cout<<endl;}
void yes(bool a){cout<<(a?"yes":"no")<<endl;}
void YES(bool a){cout<<(a?"YES":"NO")<<endl;}
void Yes(bool a){cout<<(a?"Yes":"No")<<endl;}
void possible(bool a){ cout<<(a?"possible":"impossible")<<endl; }
void Possible(bool a){ cout<<(a?"Possible":"Impossible")<<endl; }
void POSSIBLE(bool a){ cout<<(a?"POSSIBLE":"IMPOSSIBLE")<<endl; }
template<class T>auto min(const T& a){ return *min_element(all(a)); }
template<class T>auto max(const T& a){ return *max_element(all(a)); }
template<class T,class F>void print(pair<T,F> a){cout<<a.fi<<" "<<a.se<<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 (b<a) { a=b; return 1; } return 0;}
template<class T> void ifmin(T t,T u){if(t>u){cout<<-1<<endl;}else{cout<<t<<endl;}}
template<class T> void ifmax(T t,T u){if(t>u){cout<<-1<<endl;}else{cout<<t<<endl;}}
ll fastgcd(ll u,ll v){ll shl=0;while(u&&v&&u!=v){bool eu=!(u&1);bool ev=!(v&1);if(eu&&ev){++shl;u>>=1;v>>=1;}else if(eu&&!ev){u>>=1;}else if(!eu&&ev){v>>=1;}else if(u>=v){u=(u-v)>>1;}else{ll tmp=u;u=(v-u)>>1;v=tmp;}}return !u?v<<shl:u<<shl;}
ll modPow(ll a, ll n, ll mod) { if(mod==1) return 0;ll ret = 1; ll p = a % mod; while (n) { if (n & 1) ret = ret * p % mod; p = p * p % mod; n >>= 1; } return ret; }
vector<ll> divisor(ll x){ vector<ll> ans; for(ll i = 1; i * i <= x; i++){ if(x % i == 0) {ans.push_back(i); if(i*i!=x){ ans.push_back(x / ans[i]);}}}sor(ans); return ans; }
ll pop(ll x){return __builtin_popcountll(x);}
ll poplong(ll x){ll y=-1;while(x){x/=2;y++;}return y;}
P hyou(P a){ll x=fastgcd(abs(a.fi),abs(a.se));a.fi/=x;a.se/=x;if(a.se<0){a.fi*=-1;a.se*=-1;}return a;}
P Pplus(P a,P b){ return hyou({a.fi*b.se+b.fi*a.se,a.se*b.se});}
P Ptimes(P a,ll b){ return hyou({a.fi*b,a.se});}
P Ptimes(P a,P b){ return hyou({a.fi*b.fi,a.se*b.se});}
P Pminus(P a,P b){ return hyou({a.fi*b.se-b.fi*a.se,a.se*b.se});}
P Pgyaku(P a){ return hyou({a.se,a.fi});}
template<class T>
struct Sum{
  vector<T> data;
  Sum(const vector<T>& v):data(v.size()+1){
    for(ll i=0;i<v.size();i++) data[i+1]=data[i]+v[i];
  }
  T get(ll l,ll r) const {
    return data[r]-data[l];
  }
};
template<class T>
struct Sum2{
  vector<vector<T>> data;
  Sum2(const vector<vector<T>> &v):data(v.size()+1,vector<T>(v[0].size()+1)){
    for(int i=0;i<v.size();i++) for(int j=0;j<v[i].size();j++) data[i+1][j+1]=data[i][j+1]+v[i][j];
    for(int i=0;i<v.size();i++) for(int j=0;j<v[i].size();j++) data[i+1][j+1]+=data[i+1][j];
  }
  T get(ll x1,ll y1,ll x2,ll y2) const {
    return data[x2][y2]+data[x1][y1]-data[x1][y2]-data[x2][y1];
  }
};

void cincout(){
  ios::sync_with_stdio(false);
    std::cin.tie(nullptr);
  cout<< fixed << setprecision(15);
}
using mint = modint998244353;
constexpr ll MAX = 3000000;
ll fac[MAX],finv[MAX],inv[MAX];
void COMinit(){
  fac[0]=fac[1]=1;
  finv[0]=finv[1]=1;
  inv[1]=1;
  for(int i=2;i<MAX;i++){
    fac[i]=fac[i-1]*i%mod;
    inv[i]=mod-inv[mod%i]*(mod/i)%mod;
    finv[i]=finv[i-1]*inv[i]%mod;
  }
}
ll binom(ll n,ll k){
  if(n<k) return 0;
  if(n<0||k<0) return 0;
  return fac[n]*(finv[k]*finv[n-k]%mod)%mod;
}
ll HOM(ll n,ll k){
  if(n+k-1>=n-1&&n-1>=0){
  return binom(n+k-1,n-1);
  }
  else{
    return 0;
  }
}
mint op(mint a,mint b){return a+b;}
mint e(){return 0;}
int main() {
  cincout();
  COMinit();
  ll n;
  cin>>n;
  vector<ll> a(n,0);
  ll k,x;
  cin>>k>>x;
  a[k-1]=x;
  for(auto &e:a) e--;
  mint ans=0;
  ll c=0;
  for(auto e:a){
    if(e==-1) c++;
  }
  vector<ll> ok(n,1);
  for(auto e:a){
    if(e!=-1) ok[e]=0;
  }
  mint now=0;
  vector<mint> tyouka(n);
  for(int i=n-1;i>=0;i--){
    if(ok[i]) now++;
    else tyouka[i]=now;
  }
  now=0;
  vector<mint> miman(n);
  for(int i=0;i<n;i++){
    if(ok[i]) now++;
    else miman[i]=now;
  }
  mint minus=0;
  mint tyoukasum=0;
  segtree<mint,op,e> seg(n+10);
  for(int i=n-1;i>=0;i--){
    if(a[i]==-1){
      ans+=minus/mint(2)*fac[n-i-1]*fac[c];
      ans+=tyoukasum*fac[n-i-1]*fac[c-1];
      minus++;
    }
    else{
     // cout<<seg.prod(0,a[i]).val()<<" "<<fac[n-i-1]<<" "<<fac[c]<<endl;
      if(c&&i!=n-1) ans+=miman[a[i]]*fac[n-i-1]*fac[c-1]*minus;
      ans+=seg.prod(0,a[i])*fac[n-i-1]*fac[c];
      seg.set(a[i],1);
      tyoukasum+=tyouka[a[i]];
    }
    //cout<<ans.val()<<endl;
  }
  cout<<(ans+fac[c]-fac[n-1]).val()<<endl;
}
0