#1153170 (C++23) No.3478 XOR-Folding Primes

• 問題: No.3478 XOR-Folding Primes
• ユーザー: Kude
• 提出日時: 2026-03-20 22:27:03
• 言語: C++23 (gcc 15.2.0 + boost 1.89.0)
• 結果: AC
• 実行時間: 185 ms / 4,000 ms
• コード長: 3,916 bytes
• コンパイル時間: 2,207 ms
• コンパイル使用メモリ: 362,184 KB
• 実行使用メモリ: 84,984 KB
• 最終ジャッジ日時: 2026-03-20 22:27:07
• 合計ジャッジ時間: 3,739 ms
• ジャッジサーバーID （参考情報）: judge2_0 / judge1_0

ファイルパターン	結果
sample	AC * 1
other	AC * 8

ソースコード

#include<bits/stdc++.h>
namespace {
#pragma GCC diagnostic ignored "-Wunused-function"
#include<atcoder/all>
#pragma GCC diagnostic warning "-Wunused-function"
using namespace std;
using namespace atcoder;
#define rep(i,n) for(int i = 0; i < (int)(n); i++)
#define rrep(i,n) for(int i = (int)(n) - 1; i >= 0; i--)
#define all(x) begin(x), end(x)
#define rall(x) rbegin(x), rend(x)
template<class T> bool chmax(T& a, const T& b) { if (a < b) { a = b; return true; } else return false; }
template<class T> bool chmin(T& a, const T& b) { if (b < a) { a = b; return true; } else return false; }
using ll = long long;
using P = pair<int,int>;
using VI = vector<int>;
using VVI = vector<VI>;
using VL = vector<ll>;
using VVL = vector<VL>;
using mint = modint998244353;


template<class T, T (*add)(T, T), T (*zero)(), T (*mul)(T, T), T (*one)(), int N>
struct Mat : array<array<T, N>, N> {
  using M = Mat<T, add, zero, mul, one, N>;
  void make_identity() {
    for (int i = 0; i < N; i++) {
      for (int j = 0; j < N; j++) {
        (*this)[i][j] = zero();
      }
    }
    for (int i = 0; i < N; i++) {
      (*this)[i][i] = one();
    }
  }
  M& operator+=(const M& rhs) {
    for (int i = 0; i < N; i++) {
      for (int j = 0; j < N; j++) {
        (*this)[i][j] = add((*this)[i][j], rhs[i][j]);
      }
    }
    return *this;
  }
  M& operator*=(const M& rhs) {
    static M temp;
    for (int i = 0; i < N; i++) {
      for (int j = 0; j < N; j++) {
        temp[i][j] = zero();
      }
    }
    for (int i = 0; i < N; i++) {
      for (int j = 0; j < N; j++) {
        for (int k = 0; k < N; k++) {
          temp[i][k] = add(temp[i][k], mul((*this)[i][j], rhs[j][k]));
        }
      }
    }
    *this = temp;
    return *this;
  }
  template <class I>
  void inplace_pow(I k) {
    assert(k >= 0);
    static M temp;
    temp = *this;
    make_identity();
    while (k) {
      if (k & 1) *this *= temp;
      k >>= 1;
      if (k) temp *= temp;
    }
  }
  friend ostream& operator<<(ostream& os, const M& A) {
    for (int i = 0; i < N; i++) {
      for (int j = 0; j < N; j++) {
        os << A[i][j] << " \n"[j + 1 == N];
      }
    }
    return os;
  }
};

mint add(mint x, mint y) { return x + y; }
mint zero() { return mint(); }
mint mul(mint x, mint y) { return x * y; }
mint one() { return mint::raw(1); }
using M = Mat<mint, add, zero, mul, one, 2>;

pair<vector<int>, vector<int>> primes_lpf(const int n) {
  vector<int> primes; primes.reserve(n / 10);
  vector<int> lpf(n + 1);
  for (int i = 2; i <= n; i += 2) lpf[i] = 2;
  for (int i = 3; i <= n; i += 6) lpf[i] = 3;
  if (2 <= n) primes.push_back(2);
  if (3 <= n) primes.push_back(3);
  // 5 * x <= n, x <= floor(n / 5)
  const int n5 = n / 5;
  int x = 5;
  char add_next = 2;
  for (; x <= n5; x += add_next, add_next ^= 0x2 ^ 4) {
    int px = lpf[x];
    if (px == 0) {
      lpf[x] = px = x;
      primes.push_back(x);
    }
    for (int i = 2;; ++i) {
      int q = primes[i];
      int y = q * x;
      if (y > n) break;
      lpf[y] = q;
      if (q == px) break;
    }
  }
  for (; x <= n; x += add_next, add_next ^= 0x2 ^ 4) {
    if (lpf[x] == 0) {
      lpf[x] = x;
      primes.push_back(x);
    }
  }
  return {move(primes), move(lpf)};
}

constexpr int PSIZE = 10000010;
auto [primes, lpf] = primes_lpf(PSIZE);

int acc[PSIZE + 10];

} int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  rep(i, ssize(primes) - 1) {
    if ((primes[i] ^ 2) == primes[i+1]) acc[i+2]++;
  }
  rep(i, PSIZE + 9) acc[i+1] += acc[i];
  int tt;
  cin >> tt;
  while (tt--) {
    int n, m;
    cin >> n >> m;
    int k = upper_bound(all(primes), m) - primes.begin();
    if (n == 1) {
      cout << k << '\n';
      continue;
    }
    k = acc[k];
    M a;
    a[0][1] = 2 * k;
    a[1][1] = 1;
    a[1][0] = 1;
    a.inplace_pow(n - 1);
    mint ans = a[0][0] + a[0][1] + 2 * k * (a[1][0] + a[1][1]);
    cout << ans.val() << '\n';
  }
}