結果
| 問題 | No.1068 #いろいろな色 / Red and Blue and more various colors (Hard) | 
| コンテスト | |
| ユーザー |  | 
| 提出日時 | 2020-08-01 23:35:50 | 
| 言語 | C++17 (gcc 13.3.0 + boost 1.87.0) | 
| 結果 | 
                                WA
                                 
                             | 
| 実行時間 | - | 
| コード長 | 2,029 bytes | 
| コンパイル時間 | 6,999 ms | 
| コンパイル使用メモリ | 334,992 KB | 
| 最終ジャッジ日時 | 2025-01-12 13:00:00 | 
| ジャッジサーバーID (参考情報) | judge2 / judge1 | 
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| ファイルパターン | 結果 | 
|---|---|
| sample | AC * 3 | 
| other | AC * 1 WA * 10 TLE * 18 | 
ソースコード
#include <algorithm>
#include <boost/multiprecision/cpp_int.hpp>
#include <complex>
#include <iomanip>
#include <iostream>
#include <queue>
#include <set>
#include <vector>
using namespace std;
using namespace boost::multiprecision;
using ll = uint128_t;
#define rep(i, j, n) for (int i = j; i < (int)n; ++i)
constexpr ll MOD = 998244353;
namespace fft {
using C = complex<long double>;
const long double PI = 3.1415926535897932384626433832795028841971;
void fft(vector<C>& f, int n, int sgn = 1) {
  if (n == 1) return;
  vector<C> f0(n / 2), f1(n / 2);
  for (int i = 0; i < n / 2; ++i) {
    f0[i] = f[i * 2];
    f1[i] = f[i * 2 + 1];
  }
  fft(f0, n / 2, sgn);
  fft(f1, n / 2, sgn);
  C zeta = {cos(2.0 * PI / (long double)n),
            sin(2.0 * PI / (long double)n) * sgn};
  C now = {1.0, 0.0};
  for (int i = 0; i < n; ++i) {
    f[i] = f0[i % (n / 2)] + now * f1[i % (n / 2)];
    now *= zeta;
  }
}
vector<uint128_t> multiply(vector<uint128_t> a, vector<uint128_t> b) {
  int n_ = (int)a.size() + (int)b.size() - 1;
  int n = 1;
  while (n < (int)a.size() * 2 || n < (int)b.size() * 2) n <<= 1;
  vector<C> f(n), g(n);
  for (int i = 0; i < (int)a.size(); ++i) f[i] = {(long double)a[i], 0.0};
  for (int i = 0; i < (int)b.size(); ++i) g[i] = {(long double)b[i], 0.0};
  fft(f, n);
  fft(g, n);
  for (int i = 0; i < n; ++i) f[i] = f[i] * g[i];
  // inverse
  fft(f, n, -1);
  vector<uint128_t> c(n);
  for (int i = 0; i < n; ++i)
    c[i] = (uint128_t)(f[i].real() / (long double)n + 0.5) % MOD;
  c.resize(n_);
  return c;
}
} // namespace fft
vector<ll> solve(vector<ll> a) {
  int n = a.size();
  if (n == 1) return {1, a[0] - 1};
  vector<ll> f(a.begin(), a.begin() + n / 2);
  vector<ll> g(a.begin() + n / 2, a.end());
  return fft::multiply(solve(f), solve(g));
}
int main() {
  int n, q, b;
  cin >> n >> q;
  vector<ll> a(n);
  rep(i, 0, n) cin >> a[i];
  vector<ll> c = solve(a);
  reverse(c.begin(), c.end());
  rep(i, 0, q) {
    cin >> b;
    cout << c[b] % MOD << '\n';
  }
  return 0;
}
            
            
            
        