結果
問題 | No.2326 Factorial to the Power of Factorial to the... |
ユーザー |
|
提出日時 | 2023-05-28 10:22:51 |
言語 | C++17 (gcc 13.3.0 + boost 1.87.0) |
結果 |
AC
|
実行時間 | 9 ms / 2,000 ms |
コード長 | 3,838 bytes |
コンパイル時間 | 1,158 ms |
コンパイル使用メモリ | 118,292 KB |
最終ジャッジ日時 | 2025-02-13 09:32:39 |
ジャッジサーバーID (参考情報) |
judge4 / judge1 |
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ファイルパターン | 結果 |
---|---|
sample | AC * 2 |
other | AC * 20 |
ソースコード
#include <algorithm>#include <bitset>#include <cassert>#include <cmath>#include <complex>#include <cstdio>#include <deque>#include <functional>#include <iostream>#include <map>#include <numeric>#include <queue>#include <set>#include <stack>#include <string>#include <unordered_map>#include <unordered_set>#include <vector>#define REP(i, N) for (int i = 0; i < (int)N; i++)#define FOR(i, a, b) for (int i = a; i < (int)b; i++)#define ALL(x) (x).begin(), (x).end()using namespace std;constexpr int inf = 1 << 30;constexpr long long llinf = 1LL << 62;constexpr int mod = 1000000007;using ll = long long;template <int MOD = 1000000007>struct Math {vector<long long> fact, factinv, inv;Math(int n = 100000) {fact.resize(n + 1);factinv.resize(n + 1);inv.resize(n + 1);fact[0] = fact[1] = 1;factinv[0] = factinv[1] = 1;inv[1] = 1;for (int i = 2; i <= n; ++i) {fact[i] = fact[i - 1] * i % MOD;inv[i] = MOD - inv[MOD % i] * (MOD / i) % MOD;factinv[i] = factinv[i - 1] * inv[i] % MOD;}}long long C(int n, int r) {if (n < r || n < 0 || r < 0) {return 0;} else {return fact[n] * (factinv[r] * factinv[n - r] % MOD) % MOD;}}long long P(int n, int r) {if (n < r || n < 0 || r < 0) {return 0;} else {return fact[n] * factinv[n - r] % MOD;}}long long H(int n, int r) { return C(n + r - 1, r); }};namespace phc {long long modpow(long long a, long long n) {long long res = 1;while (n > 0) {if (n & 1) res = res * a % mod;a = a * a % mod;n >>= 1;}return res;}long long modinv(long long a) {long long 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);}u %= mod;if (u < 0) u += mod;return u;}long long gcd(long long a, long long b) { return b != 0 ? gcd(b, a % b) : a; }long long lcm(long long a, long long b) { return a / gcd(a, b) * b; }} // namespace phctemplate <int mod>struct ModInt {int x;ModInt() : x(0) {}ModInt(int64_t y) : x(y >= 0 ? y % mod : (mod - (-y) % mod) % mod) {}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 inverse() const {int a = x, b = mod, u = 1, v = 0, t;while (b > 0) {t = a / b;swap(a -= t * b, b);swap(u -= t * v, v);}return ModInt(u);}ModInt pow(int64_t n) const {ModInt ret(1), mul(x);while (n > 0) {if (n & 1) ret *= mul;mul *= mul;n >>= 1;}return ret;}friend ostream& operator<<(ostream& os, const ModInt& p) { return os << p.x; }friend istream& operator>>(istream& is, ModInt& a) {int64_t t;is >> t;a = ModInt<mod>(t);return (is);}static int get_mod() { return mod; }};using modint = ModInt<mod>;int main() {ll N, P;cin >> N >> P;modint ans = 0;for (ll x = 1; x <= N; ++x) {ll y = x;while (y % P == 0) {ans += 1;y /= P;}}Math<mod> m1(N);Math<mod - 1> m2(N);cout << (ans * phc::modpow(m1.fact[N], m2.fact[N])).x << endl;return 0;}