我常常追忆过去

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我常常追忆过去。
——联合省选2025 追忆

题意

请以《我常常追忆过去》为题写一篇文章，字数不限，文体不限。

思路

我常常追忆过去。

生命瞬间定格在脑海。我将背后的时间裁剪、折叠、蜷曲，揉捻成天上朵朵白云。

云朵之间亦有分别：积云厚重，而卷云飘渺。生命里震撼的场景掠过我的思绪便一生无法忘怀，而更为普通平常的记忆在时间的冲刷下只留下些许残骸。追忆宛如入梦，太过清楚则无法愉悦自己的幻想，过分模糊却又坠入虚无。只有薄雾间的山水，面纱下的女子，那恰到好处的朦胧，才能满足我对美的苛求。

追忆总在不经意间将我裹进泛黄的纸页里。分别又重聚的朋友，推倒又重建的街道，种种线索协助着我从一个具体的时刻出发沿时间的河逆流而上。曾经的日子无法重来，我只不过是一个过客。但我仍然渴望在每一次追忆之旅中留下闲暇时间，在一个场景前驻足，在岁月的朦胧里瞭望过去的自己，感受尽可能多的甜蜜。美好的时光曾流过我的身体，我便心满意足。

过去已经凝固，我带着回忆向前，只是时常疏于保管，回忆也在改变着各自的形态。这给我的追忆旅程带来些许挑战。

我该在哪里停留？我问我自己。

我常常追忆过去

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#include <bits/stdc++.h>
using namespace std;

const int N = 2e6 + 5;
const int mod = 998244353;

namespace NTT {
const long long g = 3;
const long long p = ::mod;
long long wn[35];
long long inv[N];

long long power(long long a, long long b = p - 2) {
  long long res = 1;
  while (b) {
    if (b & 1) res = res * a % p;
    a = a * a % p;
    b >>= 1;
  }
  return res;
}

int _ = [] {
  for (int i = 0; i < 35; i++) wn[i] = power(g, (p - 1) / (1ll << i));
  inv[0] = inv[1] = 1;
  for (long long i = 2; i < N; i++) inv[i] = ((p - p / i) * inv[p % i]) % p;
  return 0;
}();

void ntt(long long *a, int len, int flag) {
  long long i, j = 0, t, k, w, id;
  for (i = 1; i < len - 1; i++) {
    for (t = len; j ^= t >>= 1, ~j & t;);
    if (i < j) swap(a[i], a[j]);
  }
  for (i = 1, id = 1; i < len; i <<= 1, id++) {
    t = i << 1;
    for (j = 0; j < len; j += t) {
      for (k = 0, w = 1; k < i; k++, w = w * wn[id] % p) {
        long long x = a[j + k], y = w * a[j + k + i] % p;
        a[j + k] = (x + y) % p;
        a[j + k + i] = (x - y + p) % p;
      }
    }
  }
  if (!flag) {
    for (i = 1, j = len - 1; i < j; i++, j--) swap(a[i], a[j]);
    long long in = power(len);
    for (i = 0; i < len; i++) a[i] = a[i] * in % p;
  }
}

void mul(vector<long long> &a, vector<long long> &b) {
  int l1 = a.size(), l2 = b.size();
  int len, i;
  for (len = 1; len <= l1 + l2; len <<= 1);
  static long long A[N << 3], B[N << 3];
  a.resize(len, 0);
  b.resize(len, 0);
  for (i = 0; i < len; i++) A[i] = a[i];
  for (i = 0; i < len; i++) B[i] = b[i];
  ntt(A, len, 1);
  ntt(B, len, 1);
  for (i = 0; i < len; i++) A[i] = A[i] * B[i] % p;
  ntt(A, len, 0);
  for (i = 0; i < len; i++) a[i] = A[i];
  a.resize(l1 + l2);
  b.resize(l2, 0);
}

void invp(long long *a, int n) {
  static long long w[N << 1], r[N << 1], sav[N << 1];
  int len, k, i;
  for (len = 1; len < n; len <<= 1);
  w[0] = power(a[0]);
  for (k = 2; k <= len; k <<= 1) {
    for (i = 0; i < (k >> 1); i++) r[i] = w[i];
    for (i = 0; i < k; i++) sav[i] = a[i];
    ntt(sav, k, 1);
    ntt(r, k, 1);
    for (i = 0; i < k; i++) r[i] = r[i] * sav[i] % p;
    ntt(r, k, 0);
    memset(r, 0, sizeof(long long) * (k >> 1));
    for (i = 0; i < k; i++) sav[i] = w[i];
    ntt(sav, k, 1);
    ntt(r, k, 1);
    for (i = 0; i < k; i++) r[i] = r[i] * sav[i] % p;
    ntt(r, k, 0);
    for (i = k >> 1; i < k; i++) w[i] = (w[i] * 2 - r[i] + p) % p;
  }
  for (i = 0; i < n; i++) a[i] = w[i];
  memset(w, 0, sizeof(long long) * len);
  memset(r, 0, sizeof(long long) * len);
  memset(sav, 0, sizeof(long long) * len);
}

void invp(vector<long long> &a, int n) {
  static long long f[N << 1];
  if (a.size() < n) a.resize(n, 0);
  for (int i = 0; i < n; i++) f[i] = a[i];
  invp(f, n);
  for (int i = 0; i < n; i++) a[i] = f[i];
  a.resize(n);
  memset(f, 0, sizeof(long long) * n);
}

void deriv(vector<long long> &a) {
  if (a.empty()) return;
  for (int i = 1; i < a.size(); i++) a[i - 1] = a[i] * i % p;
  a.pop_back();
}

void integ(vector<long long> &a) {
  a.push_back(0);
  for (int i = (int)a.size() - 1; i > 0; i--) a[i] = a[i - 1] * inv[i] % p;
  a[0] = 0;
}

void sqrtp(vector<long long> &a, int n) {
  int len, k, i;
  for (len = 1; len < n; len <<= 1);
  static long long b1[N << 2], b2[N << 2], sav[N << 2];
  a.resize(len, 0);
  b1[0] = 1;
  for (k = 2; k <= len; k <<= 1) {
    for (i = 0; i < (k >> 1); i++) b2[i] = (b1[i] << 1) % p;
    invp(b2, k);
    ntt(b1, k, 1);
    for (i = 0; i < k; i++) b1[i] = b1[i] * b1[i] % p;
    ntt(b1, k, 0);
    for (i = 0; i < k; i++) b1[i] = (a[i] + b1[i]) % p;
    for (i = 0; i < k; i++) sav[i] = b2[i];
    ntt(b1, k << 1, 1);
    ntt(sav, k << 1, 1);
    for (i = 0; i < (k << 1); i++) b1[i] = b1[i] * sav[i] % p;
    ntt(b1, k << 1, 0);
    memset(b1 + k, 0, sizeof(long long) * k);
  }
  for (i = 0; i < n; i++) a[i] = b1[i];
  a.resize(n);
  memset(b1, 0, sizeof(long long) * (len << 1));
  memset(b2, 0, sizeof(long long) * (len << 1));
}

void lnp(vector<long long> &a, int n) {
  vector<long long> f;
  f = a;
  invp(f, n);
  deriv(a);
  mul(a, f);
  integ(a);
  a.resize(n, 0);
}

void exp(vector<long long> &a, int n) {
  vector<long long> s(1, 0), s2(1, 0);
  int len, k, i;
  for (len = 1; len < n; len <<= 1);
  a.resize(len, 0);
  s2[0] = 1;
  for (k = 2; k <= len; k <<= 1) {
    s.resize(k, 0);
    s2.resize(k, 0);
    for (i = 0; i < (k >> 1); i++) s[i] = s2[i];
    lnp(s, k);
    for (i = 0; i < k; i++) s[i] = (a[i] - s[i] + p) % p;
    s[0] = (s[0] + 1) % p;
    mul(s2, s);
    s2.resize(k, 0);
  }
  for (i = 0; i < n; i++) a[i] = s2[i];
  a.resize(n);
}

void cdqmul(vector<long long> &f, vector<long long> &g, int l, int r) {
  if (l >= r) return;
  int mid = (l + r) / 2;
  cdqmul(f, g, l, mid);
  vector<long long> lf, lg;
  for (int i = l; i <= mid; ++i) lf.push_back(f[i]), lg.push_back(g[i - l]);
  for (int i = mid + 1; i <= r; ++i) lg.push_back(g[i - l]);
  NTT::mul(lf, lg);
  for (int i = mid + 1; i <= r; ++i) {
    int pos = i - l;
    f[i] = (f[i] + lf[pos]) % p;
  }
  cdqmul(f, g, mid + 1, r);
}

}  // namespace NTT

int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);

  vector<long long> fac(N, 0), inv(N, 0);
  fac[0] = inv[0] = 1;
  for (int i = 1; i < N; ++i) {
    fac[i] = fac[i - 1] * i % mod;
  }
  inv[N - 1] = NTT::power(fac[N - 1]);
  for (int i = N - 2; i >= 1; --i) {
    inv[i] = inv[i + 1] * (i + 1) % mod;
  }

  int m, op;
  cin >> m >> op;

  vector<long long> f(m + 5, 0), g(m + 5, 0);
  for (int i = 0; i <= m; ++i) {
    f[i] = NTT::power(2ll, 1ll * i * (i - 1) / 2) * inv[i] % mod;
    g[i] = NTT::power(2ll, 1ll * i * (i - 1) / 2) * inv[i] % mod;
  }

  NTT::mul(f, g);
  for (int i = 1; i <= m; ++i) {
    f[i] = f[i] * fac[i] % mod;
    f[i] = f[i] * NTT::power(inv[2], 1ll * i * (i - 1) / 2) % mod;
  }

  if (op == 1) {
    for (int i = 1; i <= m; ++i) {
      cout << f[i] - 1 << "\n";
    }
  } else {
    long long ans = 0;
    for (int i = 1; i <= m; ++i) {
      ans ^= f[i] - 1;
    }
    cout << ans << "\n";
  }

  return 0;
}