#834697 (C++17) No.2197 Same Dish

• 問題: No.2197 Same Dish
• ユーザー: ineedyourlovep
• 提出日時: 2023-01-21 00:38:57
• 言語: C++17 (gcc 13.3.0 + boost 1.87.0)
• 結果: AC
• 実行時間: 85 ms / 2,000 ms
• コード長: 2,621 bytes
• コンパイル時間: 2,537 ms
• コンパイル使用メモリ: 204,976 KB
• 最終ジャッジ日時: 2025-02-10 06:03:14
• ジャッジサーバーID （参考情報）: judge3 / judge4

ファイルパターン	結果
sample	AC * 3
other	AC * 20

ソースコード

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#include <bits/stdc++.h>
#define rep(i, a, n) for(int i = a; i < (n); i++)
using namespace std;
using ll = long long;
using P = pair<int, int>;
const int INF = 1001001001;
const ll LINF = 1001002003004005006ll;
//const int mod = 1000000007;
const int mod = 998244353;
//MINT
struct mint {
  unsigned x;
  mint(): x(0) {}
  mint(ll x):x((x%mod+mod)%mod) {}
  mint operator-() const { return mint(0) - *this;}
  mint operator~() const { return mint(1) / *this;}
  mint& operator+=(const mint& a) { if((x+=a.x)>=mod) x-=mod; return *this;}
  mint& operator-=(const mint& a) { if((x+=mod-a.x)>=mod) x-=mod; return *this;}
  mint& operator*=(const mint& a) { x=(unsigned long long)x*a.x%mod; return *this;}
  mint& operator/=(const mint& a) { x=(unsigned long long)x*a.pow(mod-2).x%mod; return *this;}
  mint operator+(const mint& a) const { return mint(*this) += a;}
  mint operator-(const mint& a) const { return mint(*this) -= a;}
  mint operator*(const mint& a) const { return mint(*this) *= a;}
  mint operator/(const mint& a) const { return mint(*this) /= a;}
  mint pow(ll t) const {
    if (!t) return 1;
    mint res = pow(t>>1);
    res *= res;
    return (t&1)?res*x:res;
  }
  bool operator<(const mint& a) const { return x < a.x;}
  bool operator==(const mint& a) const { return x == a.x;}
  bool operator!=(const mint& a) const { return x != a.x;}
};
mint ex(mint x, ll t) { return x.pow(t);}
istream& operator>>(istream& i, mint& a) { unsigned long long t; i>>t; a=mint(t); return i;}
ostream& operator<<(ostream& o, const mint& a) { return o<<a.x;}
//Binary Indexed Tree (Fenwick Tree)
template<typename T>
struct BIT{
  int _n;
  vector<T> data;
  BIT(): _n(0) {};
  explicit BIT(int n): _n(n), data(n) {};
  void add(int p, T x){
    assert(0 <= p && p < _n);
    p++;
    while(p <= _n){
      data[p-1] += x;
      p += p&-p;
    }
  }
  T sum(int r){
    T s = 0;
    while(r > 0){
      s += data[r-1];
      r -= r&-r;
    }
    return s;
  }
  T sum(int l, int r){
    assert(0 <= l && l <= r && r <= _n);
    return sum(r) - sum(l);
  }
};
// POWER_MODver. N^k % MOD
ll mod_pow(ll n, ll k){
  ll res = 1;
  for(; k > 0; k >>= 1){
    if(k&1) res = (res*n)%mod;
    n = (n*n)%mod;
  }
  return res;
}
int main()
{
  ll n, k;
  cin >> n >> k;
  vector<int> l(n), r(n);
  rep(i, 0, n) cin >> l[i] >> r[i];
  mint ans = 1;
  vector<P> p(n);
  rep(i, 0, n) p[i] = {l[i], r[i]};
  sort(p.begin(), p.end());
  BIT<int> seg(200005);
  rep(i, 0, n) {
    auto [l, r] = p[i];
    ans *= max(0, (int)k-seg.sum(l+1, 200005));
    seg.add(r, 1);
  }
  cout << (mint)mod_pow(k, n) - ans << endl;
  return 0;
}
 
 
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