結果

問題 No.3187 Mingle
ユーザー MMRZ
提出日時 2025-06-25 23:24:33
言語 C++23
(gcc 13.3.0 + boost 1.87.0)
結果
AC  
実行時間 763 ms / 2,500 ms
コード長 3,954 bytes
コンパイル時間 3,385 ms
コンパイル使用メモリ 284,380 KB
実行使用メモリ 38,252 KB
最終ジャッジ日時 2025-06-25 23:24:54
合計ジャッジ時間 20,534 ms
ジャッジサーバーID
（参考情報） 		judge1 / judge4
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ファイルパターン 結果
sample AC * 3
other AC * 30
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 <bits/stdc++.h>
using namespace std;
using ll = long long;
using ull = unsigned long long;
const double pi = acos(-1);
template<class T>constexpr T inf() { return ::std::numeric_limits<T>::max(); }
template<class T>constexpr T hinf() { return inf<T>() / 2; }
template <typename T_char>T_char TL(T_char cX) { return tolower(cX); }
template <typename T_char>T_char TU(T_char cX) { return toupper(cX); }
template<class T> bool chmin(T& a,T b) { if(a > b){a = b; return true;} return false; }
template<class T> bool chmax(T& a,T b) { if(a < b){a = b; return true;} return false; }
int popcnt(unsigned long long n) { int cnt = 0; for (int i = 0; i < 64; i++)if ((n >> i) & 1)cnt++; return cnt; }
int d_sum(ll n) { int ret = 0; while (n > 0) { ret += n % 10; n /= 10; }return ret; }
int d_cnt(ll n) { int ret = 0; while (n > 0) { ret++; n /= 10; }return ret; }
ll gcd(ll a, ll b) { if (b == 0)return a; return gcd(b, a%b); };
ll lcm(ll a, ll b) { ll g = gcd(a, b); return a / g*b; };
ll MOD(ll x, ll m){return (x%m+m)%m; }
ll FLOOR(ll x, ll m) {ll r = (x%m+m)%m; return (x-r)/m; }
template<class T> using dijk = priority_queue<T, vector<T>, greater<T>>;
# define all(qpqpq)           (qpqpq).begin(),(qpqpq).end()
# define UNIQUE(wpwpw)        (wpwpw).erase(unique(all((wpwpw))),(wpwpw).end())
# define LOWER(epepe)         transform(all((epepe)),(epepe).begin(),TL<char>)
# define UPPER(rprpr)         transform(all((rprpr)),(rprpr).begin(),TU<char>)
# define rep(i,upupu)         for(ll i = 0, i##_len = (upupu);(i) < (i##_len);(i)++)
# define reps(i,opopo)        for(ll i = 1, i##_len = (opopo);(i) <= (i##_len);(i)++)
# define len(x)                ((ll)(x).size())
# define bit(n)               (1LL << (n))
# define pb push_back
# define eb emplace_back
# define exists(c, e)         ((c).find(e) != (c).end())

struct INIT{
	INIT(){
		std::ios::sync_with_stdio(false);
		std::cin.tie(0);
		cout << fixed << setprecision(20);
	}
}INIT;

namespace mmrz {
	void solve();
}

int main(){
	mmrz::solve();
}
#define debug(...) (static_cast<void>(0))

using namespace mmrz;


class dynamic_modint {
	using u64 = std::uint_fast64_t;
	static u64 Modulus;
public:
	u64 a;
	static void set_mod(u64 m) { Modulus = m; }

	u64 &value() { return a; }
	dynamic_modint(u64 x = 0) : a(x % Modulus) {}
	dynamic_modint operator+(const dynamic_modint rhs) const {
		return dynamic_modint(*this) += rhs;
	}
	dynamic_modint operator-(const dynamic_modint rhs) const {
		return dynamic_modint(*this) -= rhs;
	}
	dynamic_modint operator*(const dynamic_modint rhs) const {
		return dynamic_modint(*this) *= rhs;
	}
	dynamic_modint operator/(const dynamic_modint rhs) const {
		return dynamic_modint(*this) /= rhs;
	}
	dynamic_modint &operator+=(const dynamic_modint rhs) {
		a += rhs.a;
		if (a >= Modulus) {
			a -= Modulus;
		}
		return *this;
	}
	dynamic_modint &operator-=(const dynamic_modint rhs) {
		if (a < rhs.a) {
			a += Modulus;
		}
		a -= rhs.a;
		return *this;
	}
	dynamic_modint &operator*=(const dynamic_modint rhs) {
		a = a * rhs.a % Modulus;
		return *this;
	}
	dynamic_modint &operator/=(dynamic_modint rhs) {
		u64 exp = Modulus - 2;
		while (exp) {
			if (exp % 2) {
				*this *= rhs;
			}
			rhs *= rhs;
			exp /= 2;
		}
		return *this;
	}

	friend std::ostream& operator<<(std::ostream& os, const dynamic_modint& rhs) {
		os << rhs.a;
		return os;
	}
};

inline dynamic_modint::u64 dynamic_modint::Modulus = 1;

void SOLVE(){
	int n, p;
	cin >> n >> p;
	dynamic_modint::set_mod(p);

	vector d(n+1, vector<int>());
	reps(i, n){
		for(int j = i;j <= n;j += i){
			d[j].eb(i);
		}
	}
	
	vector<dynamic_modint> f(n+1, dynamic_modint(0));
	dynamic_modint tot = 0;
	for(int i = 3;i <= n;i++){
		dynamic_modint sum = tot;
		for(auto j : d[i])sum -= f[j];
		f[i] = (sum/i + 1)*i/(i-len(d[i]));
		tot = sum + f[i]*len(d[i]);
		for(auto j : d[i])f[j] = f[i];
	}
	cout << f[n] << '\n';
}

void mmrz::solve(){
	int t = 1;
	//cin >> t;
	while(t--)SOLVE();
}
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