結果

問題 No.1973 Divisor Sequence
コンテスト
ユーザー Chanyuh
提出日時 2022-06-27 18:22:35
言語 C++17
(gcc 13.3.0 + boost 1.87.0)
結果
AC  
実行時間 534 ms / 2,000 ms
コード長 4,167 bytes
コンパイル時間 1,378 ms
コンパイル使用メモリ 133,532 KB
最終ジャッジ日時 2025-01-30 01:02:55
ジャッジサーバーID
（参考情報） 		judge4 / judge4
このコードへのチャレンジ
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ファイルパターン 結果
sample AC * 3
other AC * 22
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 <algorithm>
#include <array>
#include <bitset>
#include <cassert>
#include <ciso646>
#include <cmath>
#include <complex>
#include <cstdio>
#include <functional>
#include <iomanip>
#include <iostream>
#include <map>
#include <queue>
#include <random>
#include <set>
#include <stack>
#include <string>
#include <tuple>
#include <unordered_map>
#include <utility>
#include <vector>
using namespace std;
typedef long long ll;
const ll mod = 1000000007;
const ll INF = (ll)1000000007 * 1000000007;
typedef pair<int, int> P;
#define rep(i, n) for (int i = 0; i < n; i++)
#define per(i, n) for (int i = n - 1; i >= 0; i--)
#define Rep(i, sta, n) for (int i = sta; i < n; i++)
#define Per(i, sta, n) for (int i = n - 1; i >= sta; i--)
typedef long double ld;
const ld eps = 1e-8;
const ld pi = acos(-1.0);
typedef pair<ll, ll> LP;
int dx[8] = {1, -1, 0, 0, 1, 1, -1, -1};
int dy[8] = {0, 0, 1, -1, 1, -1, 1, -1};
template <class T>
using max_heap = priority_queue<T>;
template <class T>
using min_heap = priority_queue<T, vector<T>, greater<>>;

template <class T>
bool chmax(T &a, const T &b) {
  if (a < b) {
    a = b;
    return 1;
  }
  return 0;
}

template <class T>
bool chmin(T &a, const T &b) {
  if (b < a) {
    a = b;
    return 1;
  }
  return 0;
}

template <typename T>
map<T, int> factorize(T x) {
  map<T, int> res;
  for (T i = 2; i * i <= x; i++) {
    while (x % i == 0) {
      x /= i;
      res[i]++;
    }
  }
  if (x != 1) res[x]++;
  return res;
}

template <int mod>
struct ModInt {
  long long x;
  static constexpr int MOD = mod;

  ModInt() : x(0) {}
  ModInt(long long y) : x(y >= 0 ? y % mod : (mod - (-y) % mod) % mod) {}

  explicit operator int() const { return x; }

  ModInt &operator+=(const ModInt &p) {
    if ((x += p.x) >= mod) x -= mod;
    return *this;
  }
  ModInt &operator-=(const ModInt &p) {
    if ((x += mod - p.x) >= mod) x -= mod;
    return *this;
  }
  ModInt &operator*=(const ModInt &p) {
    x = (int)(1LL * x * p.x % mod);
    return *this;
  }
  ModInt &operator/=(const ModInt &p) {
    *this *= p.inverse();
    return *this;
  }

  ModInt operator-() const { return ModInt(-x); }
  ModInt operator+(const ModInt &p) const { return ModInt(*this) += p; }
  ModInt operator-(const ModInt &p) const { return ModInt(*this) -= p; }
  ModInt operator*(const ModInt &p) const { return ModInt(*this) *= p; }
  ModInt operator/(const ModInt &p) const { return ModInt(*this) /= p; }
  ModInt operator%(const ModInt &p) const { return ModInt(0); }

  bool operator==(const ModInt &p) const { return x == p.x; }
  bool operator!=(const ModInt &p) const { return x != p.x; }

  ModInt inverse() const {
    int a = x, b = mod, u = 1, v = 0, t;
    while (b > 0) {
      t = a / b;
      a -= t * b;
      swap(a, b);
      u -= t * v;
      swap(u, v);
    }
    return ModInt(u);
  }

  ModInt power(long long n) const {
    ModInt ret(1), mul(x);
    while (n > 0) {
      if (n & 1) ret *= mul;
      mul *= mul;
      n >>= 1;
    }
    return ret;
  }

  ModInt power(const ModInt p) const { return ((ModInt)x).power(p.x); }

  friend ostream &operator<<(ostream &os, const ModInt<mod> &p) {
    return os << p.x;
  }
  friend istream &operator>>(istream &is, ModInt<mod> &a) {
    long long x;
    is >> x;
    a = ModInt<mod>(x);
    return (is);
  }
};

using modint = ModInt<mod>;

int n;
ll m;

modint calc(int x) {
  vector<vector<modint>> dp(n, vector<modint>(x + 1, 0));
  vector<vector<modint>> dpS(n, vector<modint>(x + 1, 0));
  rep(j, x + 1) dp[0][j] = 1;
  rep(j, x + 1) dpS[0][j] = j + 1;
  rep(i, n - 1) {
    rep(j, x + 1) {
      dp[i + 1][j] = dpS[i][x - j];
      dpS[i + 1][j] = dp[i + 1][j];
      if (j >= 1) dpS[i + 1][j] += dpS[i + 1][j - 1];
    }
  }
  // rep(i, n) { rep(j, x + 1) cout << i << " " << j << " " << dp[i][j] << endl;
  // }
  return dpS[n - 1][x];
}

void solve() {
  cin >> n >> m;
  map<ll, int> f = factorize(m);
  modint ans = 1;
  for (auto p : f) {
    // cout << p.first << " " << p.second << endl;
    ans *= calc(p.second);
  }
  cout << ans << endl;
}

int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout << fixed << setprecision(50);
  solve();
}
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