#837008 (C++14) No.2206 Popcount Sum 2

提出ソース

結果

問題	No.2206 Popcount Sum 2
ユーザー	👑 hos.lyric
提出日時	2023-02-03 21:42:13
言語	C++14 (gcc 12.3.0 + boost 1.83.0)
結果	WA
実行時間	-
コード長	5,831 bytes
コンパイル時間	1,194 ms
コンパイル使用メモリ	110,964 KB
実行使用メモリ	11,264 KB
最終ジャッジ日時	2024-07-02 19:32:47
合計ジャッジ時間	8,272 ms
ジャッジサーバーID （参考情報）	judge5 / judge4

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テストケース

テストケース表示

入力	結果	実行時間実行使用メモリ
testcase_00	AC	8 ms 8,436 KB
testcase_01	WA	-
testcase_02	WA	-
testcase_03	WA	-
testcase_04	WA	-
testcase_05	WA	-
testcase_06	WA	-
testcase_07	WA	-
testcase_08	WA	-
testcase_09	WA	-
testcase_10	WA	-
testcase_11	WA	-
testcase_12	WA	-
testcase_13	WA	-
testcase_14	WA	-
testcase_15	WA	-
testcase_16	WA	-
testcase_17	WA	-
testcase_18	WA	-

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ソースコード

raw source code

#include <cassert>
#include <cmath>
#include <cstdint>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <bitset>
#include <complex>
#include <deque>
#include <functional>
#include <iostream>
#include <map>
#include <numeric>
#include <queue>
#include <set>
#include <sstream>
#include <string>
#include <unordered_map>
#include <unordered_set>
#include <utility>
#include <vector>

using namespace std;

using Int = long long;

template <class T1, class T2> ostream &operator<<(ostream &os, const pair<T1, T2> &a) { return os << "(" << a.first << ", " << a.second << ")"; };
template <class T> ostream &operator<<(ostream &os, const vector<T> &as) { const int sz = as.size(); os << "["; for (int i = 0; i < sz; ++i) { if (i >= 256) { os << ", ..."; break; } if (i > 0) { os << ", "; } os << as[i]; } return os << "]"; }
template <class T> void pv(T a, T b) { for (T i = a; i != b; ++i) cerr << *i << " "; cerr << endl; }
template <class T> bool chmin(T &t, const T &f) { if (t > f) { t = f; return true; } return false; }
template <class T> bool chmax(T &t, const T &f) { if (t < f) { t = f; return true; } return false; }

////////////////////////////////////////////////////////////////////////////////
template <unsigned M_> struct ModInt {
  static constexpr unsigned M = M_;
  unsigned x;
  constexpr ModInt() : x(0U) {}
  constexpr ModInt(unsigned x_) : x(x_ % M) {}
  constexpr ModInt(unsigned long long x_) : x(x_ % M) {}
  constexpr ModInt(int x_) : x(((x_ %= static_cast<int>(M)) < 0) ? (x_ + static_cast<int>(M)) : x_) {}
  constexpr ModInt(long long x_) : x(((x_ %= static_cast<long long>(M)) < 0) ? (x_ + static_cast<long long>(M)) : x_) {}
  ModInt &operator+=(const ModInt &a) { x = ((x += a.x) >= M) ? (x - M) : x; return *this; }
  ModInt &operator-=(const ModInt &a) { x = ((x -= a.x) >= M) ? (x + M) : x; return *this; }
  ModInt &operator*=(const ModInt &a) { x = (static_cast<unsigned long long>(x) * a.x) % M; return *this; }
  ModInt &operator/=(const ModInt &a) { return (*this *= a.inv()); }
  ModInt pow(long long e) const {
    if (e < 0) return inv().pow(-e);
    ModInt a = *this, b = 1U; for (; e; e >>= 1) { if (e & 1) b *= a; a *= a; } return b;
  }
  ModInt inv() const {
    unsigned a = M, b = x; int y = 0, z = 1;
    for (; b; ) { const unsigned q = a / b; const unsigned c = a - q * b; a = b; b = c; const int w = y - static_cast<int>(q) * z; y = z; z = w; }
    assert(a == 1U); return ModInt(y);
  }
  ModInt operator+() const { return *this; }
  ModInt operator-() const { ModInt a; a.x = x ? (M - x) : 0U; return a; }
  ModInt operator+(const ModInt &a) const { return (ModInt(*this) += a); }
  ModInt operator-(const ModInt &a) const { return (ModInt(*this) -= a); }
  ModInt operator*(const ModInt &a) const { return (ModInt(*this) *= a); }
  ModInt operator/(const ModInt &a) const { return (ModInt(*this) /= a); }
  template <class T> friend ModInt operator+(T a, const ModInt &b) { return (ModInt(a) += b); }
  template <class T> friend ModInt operator-(T a, const ModInt &b) { return (ModInt(a) -= b); }
  template <class T> friend ModInt operator*(T a, const ModInt &b) { return (ModInt(a) *= b); }
  template <class T> friend ModInt operator/(T a, const ModInt &b) { return (ModInt(a) /= b); }
  explicit operator bool() const { return x; }
  bool operator==(const ModInt &a) const { return (x == a.x); }
  bool operator!=(const ModInt &a) const { return (x != a.x); }
  friend std::ostream &operator<<(std::ostream &os, const ModInt &a) { return os << a.x; }
};
////////////////////////////////////////////////////////////////////////////////

constexpr unsigned MO = 998244353;
using Mint = ModInt<MO>;

constexpr int LIM_INV = 400'010;
Mint inv[LIM_INV], fac[LIM_INV], invFac[LIM_INV];

void prepare() {
  inv[1] = 1;
  for (int i = 2; i < LIM_INV; ++i) {
    inv[i] = -((Mint::M / i) * inv[Mint::M % i]);
  }
  fac[0] = invFac[0] = 1;
  for (int i = 1; i < LIM_INV; ++i) {
    fac[i] = fac[i - 1] * i;
    invFac[i] = invFac[i - 1] * inv[i];
  }
}
Mint binom(Int n, Int k) {
  if (n < 0) {
    if (k >= 0) {
      return ((k & 1) ? -1 : +1) * binom(-n + k - 1, k);
    } else if (n - k >= 0) {
      return (((n - k) & 1) ? -1 : +1) * binom(-k - 1, n - k);
    } else {
      return 0;
    }
  } else {
    if (0 <= k && k <= n) {
      assert(n < LIM_INV);
      return fac[n] * invFac[k] * invFac[n - k];
    } else {
      return 0;
    }
  }
}


int T;
vector<int> N, M;

int main() {
  prepare();
  
  for (; ~scanf("%d", &T); ) {
    N.resize(T);
    M.resize(T);
    for (int t = 0; t < T; ++t) {
      scanf("%d%d", &N[t], &M[t]);
    }
    
    vector<Mint> ans(T);
    for (int t = 0; t < T; ++t) {
      ans[t] = Mint(2).pow(N[t]) - 1;
      --N[t];
      --M[t];
    }
    
    const int maxN = *max_element(N.begin(), N.end());
    const int B = max((int)(maxN / sqrt(T)), 1);
    vector<int> ts(T);
    for (int t = 0; t < T; ++t) {
      ts[t] = t;
    }
    sort(ts.begin(), ts.end(), [&](int t0, int t1) -> bool {
      const int xa = M[t0] / B;
      const int xb = M[t1] / B;
      return ((xa != xb) ? (xa < xb) : (xa & 1) ? (N[t0] < N[t1]) : (N[t0] > N[t1]));
    });
    Mint x = 1, y = 1;
    int m = 0, n = 0;
    for (int q = 0; q < T; ++q) {
      const int t = ts[q];
      for (; m > M[t]; ) {
        --m;
        x -= y;
        y *= inv[n - m] * (m + 1);
      }
      for (; n < N[t]; ) {
        ++n;
        x = 2 * x - y;
        y *= (n + 1) * inv[n - m + 1];
      }
      for (; m < M[t]; ) {
        y *= (n - m) * inv[m + 1];
        x += y;
        ++m;
      }
      for (; n > N[t]; ) {
        --n;
        y *= inv[n + 1] * (n - m + 1);
        x = inv[2] * (x + y);
      }
      ans[t] *= x;
    }
    
    for (int t = 0; t < T; ++t) {
      printf("%u\n", ans[t].x);
    }
  }
  return 0;
}

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テストケース

ソースコード