結果

問題 No.2097 AND^k
コンテスト
ユーザー chro_96
提出日時 2022-11-09 20:58:54
言語 C
(gcc 13.3.0)
結果
TLE  
実行時間 -
コード長 4,235 bytes
コンパイル時間 539 ms
コンパイル使用メモリ 34,176 KB
実行使用メモリ 163,632 KB
最終ジャッジ日時 2024-07-23 02:19:36
合計ジャッジ時間 19,293 ms
ジャッジサーバーID
（参考情報） 		judge5 / judge3
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 * 10 TLE * 1 -- * 12
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 # 

#include <stdio.h>

long long mod_num = 998244353LL;
long long root = 3LL;
int length = 998244352;
long long inverse_root = 0LL;
long long inverse_l = 0LL;

long long power_mod (long long a, long long b, long long p) {
  long long ans = 0LL;
  
  a %= p;
  
  if (b <= 0LL) {
    return 1LL;
  }
  
  ans = power_mod(a, b/2LL, p);
  ans = (ans * ans) % p;
  if (b%2LL == 1LL) {
    ans = (ans * a) % p;
  }
  
  return ans;
}

void setup_ntt (int l) {
  int tmp_length = 2;
  
  while(tmp_length < 2*l) {
    tmp_length *= 2;
  }
  
  root = power_mod(root, length / tmp_length, mod_num);
  inverse_root = power_mod(root, mod_num-2LL, mod_num);
  inverse_l = power_mod((long long) tmp_length, mod_num-2LL, mod_num);
  length = tmp_length;
  
  return;
}

void ntt_2n (long long *a, long long root) {
  int log = 0;
  long long pow_root[32] = {};
  
  while ((1<<log) < length) {
    log++;
  }
  
  pow_root[log-1] = root;
  for (int i = log-1; i > 0; i--) {
    pow_root[i-1] = pow_root[i];
    pow_root[i-1] *= pow_root[i];
    pow_root[i-1] %= mod_num;
  }
  
  for (int i = 0; i < length; i++) {
    int idx = 0;
    int tmp = i;
    for (int j = 0; j < log; j++) {
      idx <<= 1;
      idx |= (tmp&1);
      tmp >>= 1;
    }
    if (i < idx) {
      int swap = a[i];
      a[i] = a[idx];
      a[idx] = swap;
    }
  }
  
  for (int i = 0; i < log; i++) {
    int step = (1<<i);
    int cnt = length/(2*step);
    long long tmp_root = 1LL;
    for (int j = 0; j < step; j++) {
      for (int k = 0; k < cnt; k++) {
        long long tmp1 = a[(k<<(i+1))+j];
        long long tmp2 = (a[((2*k+1)<<i)+j]*tmp_root)%mod_num;
        a[(k<<(i+1))+j] = (tmp1+tmp2)%mod_num;
        a[((2*k+1)<<i)+j] = (tmp1+mod_num-tmp2)%mod_num;
      }
      tmp_root = (tmp_root*pow_root[i])%mod_num;
    }
  }
  
  return;
}

int main () {
  int n = 0;
  int m = 0;
  int l = 0;
  
  int res = 0;
  
  long long ans[300000] = {};
  long long work[300000] = {};
  
  long long pow2n = 0LL;
  long long pow2[100001] = {};
  long long fact[100001] = {};
  long long invf[100001] = {};
  
  long long ans_pow2[32][300000] = {};
  long long ans_pow2_dft[32][300000] = {};
  
  res = scanf("%d", &n);
  res = scanf("%d", &m);
  res = scanf("%d", &l);
  
  pow2n = power_mod(2LL, (long long)n, mod_num);

  pow2[0] = 1LL;
  for (int i = 0; i < m; i++) {
    pow2[i+1] = (pow2[i]*2LL)%mod_num;
  }
  
  fact[0] = 1LL;
  for (int i = 0; i < l; i++) {
    fact[i+1] = fact[i];
    fact[i+1] *= (long long) (i+1);
    fact[i+1] %= mod_num;
  }
  
  invf[l] = power_mod(fact[l], mod_num-2LL, mod_num);
  for (int i = l; i > 0; i--) {
    invf[i-1] = invf[i];
    invf[i-1] *= (long long)i;
    invf[i-1] %= mod_num;
  }
  
  setup_ntt(l+1);
  
  ans_pow2[0][0] = pow2n;
  for (int i = 1; i <= l; i++) {
    ans_pow2[0][i] = invf[i];
  }
  
  for (int i = 0; i <= l; i++) {
    ans_pow2_dft[0][i] = ans_pow2[0][i];
  }
  ntt_2n(ans_pow2_dft[0], root);
  
  for (int i = 1; (1<<i) <= m; i++) {
    long long p = 1LL;
    for (int j = 0; j <= l; j++) {
      ans_pow2[i][j] = ans_pow2[i-1][j];
      ans_pow2[i][j] *= p;
      ans_pow2[i][j] %= mod_num;
      p = (p*pow2[1<<(i-1)])%mod_num;
    }
    ntt_2n(ans_pow2[i], root);
    for (int j = 0; j < length; j++) {
      ans_pow2[i][j] *= ans_pow2_dft[i-1][j];
      ans_pow2[i][j] %= mod_num;
    }
    ntt_2n(ans_pow2[i], inverse_root);
    for (int j = 0; j <= l; j++) {
      ans_pow2[i][j] *= inverse_l;
      ans_pow2[i][j] %= mod_num;
      ans_pow2_dft[i][j] = ans_pow2[i][j];
    }
    ntt_2n(ans_pow2_dft[i], root);
  }
  
  ans[0] = 1LL;
  for (int i = 0; (1<<i) <= m; i++) {
    if ((m&(1<<i)) > 0) {
      long long p = 1LL;
      for (int j = 0; j <= l; j++) {
        ans[j] = (ans[j]*p)%mod_num;
        p = (p*pow2[1<<i])%mod_num;
      }
      ntt_2n(ans, root);
      for (int j = 0; j < length; j++) {
        ans[j] = (ans[j]*ans_pow2_dft[i][j])%mod_num;
      }
      ntt_2n(ans, inverse_root);
      for (int i = 0; i <= l; i++) {
        ans[i] *= inverse_l;
        ans[i] %= mod_num;
      }
      for (int i = l+1; i < length; i++) {
        ans[i] = 0LL;
      }
    }
  }
  
  for (int i = 0; i < l; i++) {
    printf("%lld\n", (ans[i+1]*fact[i+1])%mod_num);
  }
  
  return 0;
}
0