QOJ.ac

QOJ

ID题目提交者结果用时内存语言文件大小提交时间测评时间
#387464#8047. DFS Order 4wsyearTL 3ms6168kbC++143.0kb2024-04-12 15:29:552024-04-12 15:29:55

Judging History

你现在查看的是最新测评结果

  • [2024-04-12 15:29:55]
  • 评测
  • 测评结果:TL
  • 用时:3ms
  • 内存:6168kb
  • [2024-04-12 15:29:55]
  • 提交

answer

// Author: Klay Thompson
// Problem: # 8047. DFS Order 4
// Memory Limit: 1024 MB
// Time Limit: 1000 ms
// 
// Powered by CP Editor (https://cpeditor.org)

#include <bits/stdc++.h>

#define rep(i, j, k) for (int i = (j); i <= (k); ++i)
#define per(i, j, k) for (int i = (j); i >= (k); --i)
#define SZ(v) int((v).size())
#define ALL(v) (v).begin(),(v).end()
#define fi first
#define se second
using ll = long long;
using pii = std::pair<int, int>;
using pll = std::pair<ll, ll>;

template<class T>inline void chkmn(T &x, T y) { if (y < x) x = y; }
template<class T>inline void chkmx(T &x, T y) { if (y > x) x = y; }

using namespace std;

int P;
class mod_int {
  using Z = mod_int;

private:
  static int mo(int x) { return x < 0 ? x + P : x; }

public:
  int x;
  int val() const { return x; }
  mod_int() : x(0) {}
  template <class T>
  mod_int(const T &x_) : x(x_ >= 0 && x_ < P ? static_cast<int>(x_) : mo(static_cast<int>(x_ % P))) {}
  bool operator==(const Z &rhs) const { return x == rhs.x; }
  bool operator!=(const Z &rhs) const { return x != rhs.x; }
  Z operator-() const { return Z(x ? P - x : 0); }
  Z pow(long long k) const {
    Z res = 1, t = *this;
    while (k) {
      if (k & 1) res *= t;
      if (k >>= 1) t *= t;
    }
    return res;
  }
  Z &operator++() {
    x < P - 1 ? ++x : x = 0;
    return *this;
  }
  Z &operator--() {
    x ? --x : x = P - 1;
    return *this;
  }
  Z operator++(int) {
    Z ret = x;
    x < P - 1 ? ++x : x = 0;
    return ret;
  }
  Z operator--(int) {
    Z ret = x;
    x ? --x : x = P - 1;
    return ret;
  }
  Z inv() const { return pow(P - 2); }
  Z &operator+=(const Z &rhs) {
    (x += rhs.x) >= P && (x -= P);
    return *this;
  }
  Z &operator-=(const Z &rhs) {
    (x -= rhs.x) < 0 && (x += P);
    return *this;
  }
  Z &operator*=(const Z &rhs) {
    x = 1ULL * x * rhs.x % P;
    return *this;
  }
  Z &operator/=(const Z &rhs) { return *this *= rhs.inv(); }
#define setO(T, o)                                 \
  friend T operator o(const Z &lhs, const Z &rhs) {\
    Z res = lhs;                                   \
    return res o## = rhs;                          \
  }
  setO(Z, +) setO(Z, -) setO(Z, *) setO(Z, /)
#undef setO
};
using Z = mod_int;

const int maxn = 810;

int n;
Z f[maxn][maxn], fac[maxn], ivf[maxn], inv[maxn];

int main() {
  cin.tie(nullptr) -> ios::sync_with_stdio(false);
  cin >> n >> P;
  fac[0] = 1;
  rep (i, 1, n) fac[i] = fac[i - 1] * i;
  ivf[n] = fac[n].inv();
  per (i, n, 1) ivf[i - 1] = ivf[i] * i;
  rep (i, 1, n) inv[i] = fac[i - 1] * ivf[i];
  f[1][0] = 1;
  rep (i, 2, n) rep (j, 0, n) {
    f[i][j] = f[i - 1][0] * inv[i + j - 1];
    rep (k, 2, i - 1) {
      f[i][j] += f[k][0] * f[i - k][j] * inv[i + j - 1];
      f[i][j] -= f[k][i - 1 - k + j] * f[i - k][j] * inv[i + j - 1];
    }
    // cerr << i << " " << j << ": " << f[i][j].val() << '\n';
  }
  cout << (f[n][0] * fac[n - 1]).val() << '\n';
}

详细

Test #1:

score: 100
Accepted
time: 0ms
memory: 6156kb

input:

4 114514199

output:

2

result:

ok 1 number(s): "2"

Test #2:

score: 0
Accepted
time: 1ms
memory: 6156kb

input:

10 998244353

output:

11033

result:

ok 1 number(s): "11033"

Test #3:

score: 0
Accepted
time: 3ms
memory: 6168kb

input:

100 1000000007

output:

270904395

result:

ok 1 number(s): "270904395"

Test #4:

score: -100
Time Limit Exceeded

input:

756 1001338769

output:


result: