結果

問題 No.2705 L to R Graph (Another ver.)
ユーザー Magentor
提出日時 2024-03-16 01:33:33
言語 C++23
(gcc 13.3.0 + boost 1.87.0)
結果
AC  
実行時間 1,149 ms / 3,000 ms
コード長 7,979 bytes
コンパイル時間 5,918 ms
コンパイル使用メモリ 325,900 KB
実行使用メモリ 196,112 KB
最終ジャッジ日時 2024-09-30 03:40:39
合計ジャッジ時間 46,939 ms
ジャッジサーバーID
（参考情報） 		judge5 / judge4
このコードへのチャレンジ
（要ログイン）
ファイルパターン 結果
sample AC * 3
other AC * 50
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 #
プレゼンテーションモードにする

#include <bits/stdc++.h>
using namespace std;
#include <atcoder/all>
using namespace atcoder;
template<typename T> inline bool chmax(T &a, T b) { return ((a < b) ? (a = b, true) : (false)); }
template<typename T> inline bool chmin(T &a, T b) { return ((a > b) ? (a = b, true) : (false)); }
#define rep(i, n) for (long long i = 0; i < (long long)(n); i++)
#define rep2(i, m ,n) for (int i = (m); i < (long long)(n); i++)
#define REP(i, n) for (long long i = 1; i < (long long)(n); i++)
typedef long long ll;
#pragma GCC target("avx512f")
#pragma GCC optimize("Ofast")
#pragma GCC optimize("unroll-loops")
#define updiv(N,X) (N + X - 1) / X
#define l(n) n.begin(),n.end()
#define mat vector<vector<ll>>
#define YesNo(Q) Q==1?cout<<"Yes":cout<<"No"
#define int long long
using Pt = pair<int, int>;
using mint = modint;
const int MOD = 998244353LL;
const ll INF = 999999999999LL;
vector<long long> fact, fact_inv, inv;
/*  init_nCk :二項係数のための前処理
    計算量:O(n)
*/
template <typename T>
void input(vector<T> &v){
 rep(i,v.size()){cin>>v[i];}
  return;
}
void init_nCk(int SIZE) {
    fact.resize(SIZE + 5);
    fact_inv.resize(SIZE + 5);
    inv.resize(SIZE + 5);
    fact[0] = fact[1] = 1;
    fact_inv[0] = fact_inv[1] = 1;
    inv[1] = 1;
    for (int i = 2; i < SIZE + 5; i++) {
        fact[i] = fact[i - 1] * i % MOD;
        inv[i] = MOD - inv[MOD % i] * (MOD / i) % MOD;
        fact_inv[i] = fact_inv[i - 1] * inv[i] % MOD;
    }
}
/*  nCk :MODでの二項係数を求める(前処理 int_nCk が必要)
    計算量:O(1)
*/
long long nCk(int n, int k) {
    assert(!(n < k));
    assert(!(n < 0 || k < 0));
    return fact[n] * (fact_inv[k] * fact_inv[n - k] % MOD) % MOD;
}
long long modpow(long long a, long long n, long long mod) {
    long long res = 1;
    while (n > 0) {
        if (n & 1) res = res * a % mod;
        a = a * a % mod;
        n >>= 1;
    }
    return res;
}
ll POW(ll a,ll n){
  long long res = 1;
    while (n > 0) {
        if (n & 1) res = res * a;
        a = a * a;
        n >>= 1;
    }
    return res;
}
struct unionfind{
  vector<int> par,siz;
  void reset(int n){par.resize(n);siz.resize(n);rep(i,n){par[i]=-1;siz[i]=1;}}
  
  int root(int x){
   if(par[x]==-1){return x;}
   else{return par[x] = root(par[x]);} 
  }
  
  bool issame(int x,int y){
   return root(x)==root(y); 
  }
  
  bool unite(int x,int y){
   x = root(x);y=root(y);
   if(x == y){return false;}
   if(siz[x] < siz[y]){swap(x,y);} 
   par[y] = x;
  siz[x] += siz[y];
  return true; 
  }
  
  
 
int size(int x){
 return siz[root(x)]; 
}
 
};
 
  struct graph{
vector<vector< pair<int,ll> > > val;  
void print(){
 rep(i,val.size()){
  rep(j,val[i].size()){
   cout << val[i][j].first<<"/" <<val[i][j].second << " ";
  }
   cout << endl;
 }
}
void resize(int n){
 val.resize(n); 
}
void add(int n,int k,ll cost){ assert((int)(val.size())>k);val[ n ].push_back( pair(k,cost) ); }
void add2(int n,int k,ll cost){ val[ n ].push_back( pair(k,cost) ); val[ k ].push_back( pair(n,cost) );}    
vector<ll> dfs_basic(int a){
 vector<ll>seen(val.size(),-1);
 queue<int> q;q.push(a);seen[a]=0;
 
 while(!q.empty()){
  int wc=q.front();
     q.pop();
  
  rep(i,val[wc].size()){
   if(-1==seen[val[wc][i].first]){q.push(val[wc][i].first);seen[val[wc][i].first]=seen[wc]+val[wc][i].second;} 
  }
 }
 return seen;  
 }
 
    
  
};
// N の約数をすべて求める関数
ll cd(long long N) {
    // 答えを表す集合
    long long res=0;
    // 各整数 i が N の約数かどうかを調べる
    for (long long i = 1; i * i <= N; ++i) {
        // i が N の約数でない場合はスキップ
        if (N % i != 0) continue;
        // i は約数である
        res ++;
        // N ÷ i も約数である (重複に注意)
        if (N / i != i){res += 1;}
    }
    // 約数を小さい順に並び替えて出力
    
    return res;
}
ll mdt;
/// 行列積
mat mat_mul(mat &a, mat &b) {
  mat res(a.size(), vector<ll>(b[0].size()));
  for (int i = 0; i < (int)(a.size()); i++) {
    for (int j = 0; j < (int)(b[0].size()); j++) {
      for (int k = 0; k < (int)(b.size()); k++) {
        (res[i][j] += a[i][k] * b[k][j]) %= mdt;
      }
    }
  }
  return res;
}
 
/// 行列累乗
mat mat_pow(mat a, long long n) {
  mat res(a.size(), vector<ll>(a.size()));
  // 単位行列で初期化
  for (int i = 0; i < (int)(a.size()); i++)
    res[i][i] = 1;
  // 繰り返し二乗法
  while (n > 0) {
    if (n & 1) res = mat_mul(a, res);
    a = mat_mul(a, a);
    n >>= 1;
  }
  return res;
}
int N,P;
void cumsum(std::vector<int>& x, int d = 1) {
    for (int i = d; i < x.size(); i++) {
        x[i] += x[i - d];
        x[i] %= P;
    }
}
void md( long long& y ){
  y = (y%P+P)%P;
  return;
}
signed main() {
  
    std::cin >> N >> P;
    int ans = 0;
    std::vector<int> f = {0, 0};
    int fact = 1;
    for (int i = 2; i <= N; i++) {
        f.push_back((modpow(N, N - i, P) * fact) % P);
        fact *= N - i;
        fact %= P;
    }
    std::vector<int> dist(N, 0);
    int cumt = 0;
    int pat = 0;
    for (int i = N - 2; i > 0; i--) {
        cumt += f[i + 2];
        pat += cumt;
        cumt %= P;
        pat %= P;
        dist[i] = pat;
    }
    int backet = 100;
    std::vector<int> pats(N * 2, 0);
    std::vector<std::vector<int>> det(backet + 1, std::vector<int>(N * 2, 0));
    for (int i = 1; i <= N; i++) {
        for (int j = 0; j <= N; j++) {
            if (i * j >= N) {
                break;
            }
            if (j == 0) {
                pats[1] += N - i + 1;
                pats[i * (j + 1)] -= N - i + 1;
                continue;
            }
            if (j > backet) {
                int cnt = 0;
                for (int k = i * j + 1; k < i * (j + 1); k += j) {
                    pats[k] += 1;
                    cnt += 1;
                }
                pats[i * (j + 1)] -= cnt;
            } else {
                det[j][i * j + 1] += 1;
                int cnt = (i * (j + 1) - 2) / j - i + 1;
                det[j][std::min(N * 2 - 1, i * j + 1 + cnt * j)] -= 1;
                pats[i * (j + 1)] -= cnt;
            }
        }
    }
    for (int i = 1; i <= backet; i++) {
        cumsum(det[i], i);
        for (int j = 0; j < N * 2; j++) {
            pats[j] += det[i][j];
            pats[j] %= P;
        }
    }
    cumsum(pats);
    for (int i = 0; i < N - 1; i++) {
        pats[i] = N * (N + 1) / 2 - pats[i];
        pats[i] %= P;dist[i] %= P;
        ans += pats[i] * dist[i];
        ans %= P;
    }
    cumsum(f);
    std::vector<std::vector<int>> div(N + 1, std::vector<int>());
    std::vector<std::vector<int>> cum(N + 1, std::vector<int>(1, 0));
    for (int i = 1; i <= N; i++) {
        for (int j = i; j <= N; j += i) {
            div[j].push_back(i);
            cum[i].push_back(cum[i].back() + f[j]);
        }
    }
    for (int i = 1; i <= N; i++) {
        std::vector<int> gcd(div[i].size(), 0);
        for (int j = div[i].size() - 1; j >= 0; j--) {
            gcd[j] += N / div[i][j];
            for (int k = j - 1; k >= 0; k--) {
                if (div[i][j] % div[i][k] == 0) {
                    gcd[k] -= gcd[j];
                }
            }
            int pl = cum[div[i][j]].back() - cum[div[i][j]][i / div[i][j]];
            md(pl);
            int plcnt = cum[div[i][j]].size() - 1 - i / div[i][j];
            md(plcnt);
            int mi = cum[div[i][j]].back() - cum[div[i][j]][i / div[i][j] - 1] + (f[i - 1] * (i / div[i][j] - 1))%P;
            md(mi);
            int micnt = plcnt + 1 + i / div[i][j] - 1;
            md(micnt);
            ans += ((pl + f.back() * (micnt - plcnt) - mi)%P) * gcd[j];
            md(ans);
            ans %= P;
            if (j == 0) {
                ans += (((pl + f.back() * (micnt - plcnt) - mi)%P+P)%P) * ((N * (N - 1) / 2)%P);
            }
            md(ans);
        }
    }
    std::cout << (ans * ((N * (N - 1))%P)) % P << std::endl;
    return 0;
}
הההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההה
XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX
0