#684626 (C++17) No.1631 Sorting Integers (Multiple of K) Easy

提出ソース

結果

問題	No.1631 Sorting Integers (Multiple of K) Easy
ユーザー	stoq
提出日時	2021-07-30 22:19:49
言語	C++17 (gcc 12.3.0 + boost 1.83.0)
結果	WA
実行時間	-
コード長	6,908 bytes
コンパイル時間	2,235 ms
コンパイル使用メモリ	209,852 KB
実行使用メモリ	139,776 KB
最終ジャッジ日時	2024-09-16 00:45:53
合計ジャッジ時間	10,925 ms
ジャッジサーバーID （参考情報）	judge3 / judge5

このコードへのチャレンジ
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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	21 ms 11,776 KB
testcase_01	AC	22 ms 11,904 KB
testcase_02	AC	22 ms 11,904 KB
testcase_03	AC	23 ms 14,336 KB
testcase_04	AC	21 ms 11,776 KB
testcase_05	AC	21 ms 11,776 KB
testcase_06	AC	21 ms 11,904 KB
testcase_07	AC	21 ms 11,904 KB
testcase_08	AC	21 ms 11,904 KB
testcase_09	AC	22 ms 12,800 KB
testcase_10	AC	21 ms 12,416 KB
testcase_11	AC	22 ms 12,672 KB
testcase_12	AC	22 ms 12,416 KB
testcase_13	AC	22 ms 12,288 KB
testcase_14	AC	454 ms 139,776 KB
testcase_15	AC	468 ms 139,776 KB
testcase_16	AC	469 ms 139,776 KB
testcase_17	AC	485 ms 139,776 KB
testcase_18	AC	487 ms 139,776 KB
testcase_19	AC	486 ms 139,648 KB
testcase_20	AC	505 ms 139,776 KB
testcase_21	AC	451 ms 139,776 KB
testcase_22	AC	495 ms 139,776 KB
testcase_23	AC	498 ms 139,776 KB
testcase_24	AC	499 ms 139,776 KB
testcase_25	AC	476 ms 139,776 KB
testcase_26	WA	-
testcase_27	AC	391 ms 139,776 KB
testcase_28	AC	225 ms 115,200 KB
testcase_29	AC	222 ms 115,200 KB
testcase_30	AC	286 ms 134,528 KB
testcase_31	AC	386 ms 139,776 KB

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ソースコード

raw source code

#define MOD_TYPE 1

#pragma region Macros

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

#if 1
#pragma GCC target("avx2")
#pragma GCC optimize("O3")
#pragma GCC optimize("unroll-loops")
#endif

#if 0
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/pb_ds/tag_and_trait.hpp>
#include <ext/rope>
using namespace __gnu_pbds;
using namespace __gnu_cxx;
template <typename T>
using extset = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
#endif

#if 0
#include <boost/multiprecision/cpp_int.hpp>
#include <boost/multiprecision/cpp_dec_float.hpp>
using Int = boost::multiprecision::cpp_int;
using lld = boost::multiprecision::cpp_dec_float_100;
#endif

using ll = long long;
using ld = long double;
using pii = pair<int, int>;
using pll = pair<ll, ll>;
using pld = pair<ld, ld>;
template <typename T>
using smaller_queue = priority_queue<T, vector<T>, greater<T>>;

constexpr ll MOD = (MOD_TYPE == 1 ? (ll)(1e9 + 7) : 998244353);
constexpr int INF = (int)1e9 + 10;
constexpr ll LINF = (ll)4e18;
constexpr ld PI = acos(-1.0);
constexpr ld EPS = 1e-7;
constexpr int Dx[] = {0, 0, -1, 1, -1, 1, -1, 1, 0};
constexpr int Dy[] = {1, -1, 0, 0, -1, -1, 1, 1, 0};

#define REP(i, m, n) for (ll i = m; i < (ll)(n); ++i)
#define rep(i, n) REP(i, 0, n)
#define REPI(i, m, n) for (int i = m; i < (int)(n); ++i)
#define repi(i, n) REPI(i, 0, n)
#define MP make_pair
#define MT make_tuple
#define YES(n) cout << ((n) ? "YES" : "NO") << "\n"
#define Yes(n) cout << ((n) ? "Yes" : "No") << "\n"
#define possible(n) cout << ((n) ? "possible" : "impossible") << "\n"
#define Possible(n) cout << ((n) ? "Possible" : "Impossible") << "\n"
#define all(v) v.begin(), v.end()
#define NP(v) next_permutation(all(v))
#define dbg(x) cerr << #x << ":" << x << "\n";

struct io_init
{
  io_init()
  {
    cin.tie(0);
    ios::sync_with_stdio(false);
    cout << setprecision(30) << setiosflags(ios::fixed);
  };
} io_init;
template <typename T>
inline bool chmin(T &a, T b)
{
  if (a > b)
  {
    a = b;
    return true;
  }
  return false;
}
template <typename T>
inline bool chmax(T &a, T b)
{
  if (a < b)
  {
    a = b;
    return true;
  }
  return false;
}
inline ll CEIL(ll a, ll b)
{
  return (a + b - 1) / b;
}
template <typename A, size_t N, typename T>
inline void Fill(A (&array)[N], const T &val)
{
  fill((T *)array, (T *)(array + N), val);
}
template <typename T, typename U>
constexpr istream &operator>>(istream &is, pair<T, U> &p) noexcept
{
  is >> p.first >> p.second;
  return is;
}
template <typename T, typename U>
constexpr ostream &operator<<(ostream &os, pair<T, U> &p) noexcept
{
  os << p.first << " " << p.second;
  return os;
}
#pragma endregion

// --------------------------------------

#pragma region mint
template <int MOD>
struct Fp
{
  long long val;

  constexpr Fp(long long v = 0) noexcept : val(v % MOD)
  {
    if (val < 0)
      v += MOD;
  }

  constexpr int getmod()
  {
    return MOD;
  }

  constexpr Fp operator-() const noexcept
  {
    return val ? MOD - val : 0;
  }

  constexpr Fp operator+(const Fp &r) const noexcept
  {
    return Fp(*this) += r;
  }

  constexpr Fp operator-(const Fp &r) const noexcept
  {
    return Fp(*this) -= r;
  }

  constexpr Fp operator*(const Fp &r) const noexcept
  {
    return Fp(*this) *= r;
  }

  constexpr Fp operator/(const Fp &r) const noexcept
  {
    return Fp(*this) /= r;
  }

  constexpr Fp &operator+=(const Fp &r) noexcept
  {
    val += r.val;
    if (val >= MOD)
      val -= MOD;
    return *this;
  }

  constexpr Fp &operator-=(const Fp &r) noexcept
  {
    val -= r.val;
    if (val < 0)
      val += MOD;
    return *this;
  }

  constexpr Fp &operator*=(const Fp &r) noexcept
  {
    val = val * r.val % MOD;
    if (val < 0)
      val += MOD;
    return *this;
  }

  constexpr Fp &operator/=(const Fp &r) noexcept
  {
    long long a = r.val, b = MOD, u = 1, v = 0;
    while (b)
    {
      long long t = a / b;
      a -= t * b;
      swap(a, b);
      u -= t * v;
      swap(u, v);
    }
    val = val * u % MOD;
    if (val < 0)
      val += MOD;
    return *this;
  }

  constexpr bool operator==(const Fp &r) const noexcept
  {
    return this->val == r.val;
  }

  constexpr bool operator!=(const Fp &r) const noexcept
  {
    return this->val != r.val;
  }

  friend constexpr ostream &operator<<(ostream &os, const Fp<MOD> &x) noexcept
  {
    return os << x.val;
  }

  friend constexpr istream &operator>>(istream &is, Fp<MOD> &x) noexcept
  {
    return is >> x.val;
  }
};

Fp<MOD> modpow(const Fp<MOD> &a, long long n) noexcept
{
  if (n == 0)
    return 1;
  auto t = modpow(a, n / 2);
  t = t * t;
  if (n & 1)
    t = t * a;
  return t;
}

using mint = Fp<MOD>;

template <class T>
struct BiCoef
{
  vector<T> fact_, inv_, finv_;

  constexpr BiCoef()
  {
  }

  constexpr BiCoef(int n) noexcept : fact_(n, 1), inv_(n, 1), finv_(n, 1)
  {
    init(n);
  }

  constexpr void init(int n) noexcept
  {
    fact_.assign(n, 1), inv_.assign(n, 1), finv_.assign(n, 1);
    int MOD = fact_[0].getmod();
    for (int i = 2; i < n; i++)
    {
      fact_[i] = fact_[i - 1] * i;
      inv_[i] = -inv_[MOD % i] * (MOD / i);
      finv_[i] = finv_[i - 1] * inv_[i];
    }
  }

  constexpr T C(ll n, ll k) const noexcept
  {
    if (n < k || n < 0 || k < 0)
      return 0;
    return fact_[n] * finv_[k] * finv_[n - k];
  }

  constexpr T P(ll n, ll k) const noexcept
  {
    return C(n, k) * fact_[k];
  }

  constexpr T H(ll n, ll k) const noexcept
  {
    return C(n + k - 1, k);
  }

  constexpr T Ch1(ll n, ll k) const noexcept
  {
    if (n < 0 || k < 0)
      return 0;
    T res = 0;
    for (int i = 0; i < n; i++)
      res += C(n, i) * modpow(n - i, k) * (i & 1 ? -1 : 1);
    return res;
  }

  constexpr T fact(ll n) const noexcept
  {
    if (n < 0)
      return 0;
    return fact_[n];
  }

  constexpr T inv(ll n) const noexcept
  {
    if (n < 0)
      return 0;
    return inv_[n];
  }

  constexpr T finv(ll n) const noexcept
  {
    if (n < 0)
      return 0;
    return finv_[n];
  }
};

BiCoef<mint> bc(200010);
#pragma endregion

ll dp[1 << 14][1000];
int rem[1000][1000];

void solve()
{
  int n, k;
  cin >> n >> k;
  rep(i, 1000) rep(j, 1000) rem[i][j] = i * j % k;
  vector<int> v;
  int a[9];
  rep(i, 9)
  {
    cin >> a[i];
    rep(j, a[i]) v.push_back(i + 1);
  }
  dp[0][0] = 1;
  int p = 1;
  int tmp;
  repi(i, n)
  {
    repi(msk, 1 << n)
    {
      if (__builtin_popcount(msk) != i)
        continue;
      repi(r, k)
      {
        repi(j, n)
        {
          if (msk & (1 << j))
            continue;
          tmp = r + rem[v[j]][p];
          if (tmp >= k)
            tmp -= k;
          dp[msk | (1 << j)][tmp] += dp[msk][r];
        }
      }
    }
    p = rem[p][10];
  }
  mint ans = dp[(1 << n) - 1][0];
  rep(i, 9)
  {
    ans *= bc.finv(a[i]);
  }
  cout << ans << "\n";
}

int main()
{
  solve();
}

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結果

テストケース

ソースコード