結果

問題 No.2883 K-powered Sum of Fibonacci
ユーザー 👑 binapbinap
提出日時 2024-09-25 04:36:20
言語 C++17
(gcc 12.3.0 + boost 1.83.0)
結果
AC  
実行時間 3 ms / 3,000 ms
コード長 4,869 bytes
コンパイル時間 4,723 ms
コンパイル使用メモリ 270,408 KB
実行使用メモリ 6,948 KB
最終ジャッジ日時 2024-09-25 04:36:27
合計ジャッジ時間 6,235 ms
ジャッジサーバーID
(参考情報)
judge4 / judge2
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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 2 ms
6,816 KB
testcase_01 AC 2 ms
6,816 KB
testcase_02 AC 2 ms
6,812 KB
testcase_03 AC 2 ms
6,940 KB
testcase_04 AC 2 ms
6,940 KB
testcase_05 AC 2 ms
6,940 KB
testcase_06 AC 2 ms
6,940 KB
testcase_07 AC 2 ms
6,944 KB
testcase_08 AC 2 ms
6,944 KB
testcase_09 AC 2 ms
6,944 KB
testcase_10 AC 2 ms
6,944 KB
testcase_11 AC 2 ms
6,944 KB
testcase_12 AC 2 ms
6,940 KB
testcase_13 AC 2 ms
6,940 KB
testcase_14 AC 2 ms
6,944 KB
testcase_15 AC 2 ms
6,940 KB
testcase_16 AC 2 ms
6,948 KB
testcase_17 AC 3 ms
6,940 KB
testcase_18 AC 2 ms
6,940 KB
testcase_19 AC 2 ms
6,944 KB
testcase_20 AC 3 ms
6,944 KB
testcase_21 AC 2 ms
6,940 KB
testcase_22 AC 2 ms
6,944 KB
testcase_23 AC 2 ms
6,944 KB
testcase_24 AC 2 ms
6,944 KB
testcase_25 AC 2 ms
6,940 KB
testcase_26 AC 2 ms
6,944 KB
testcase_27 AC 2 ms
6,940 KB
testcase_28 AC 3 ms
6,944 KB
testcase_29 AC 2 ms
6,940 KB
testcase_30 AC 2 ms
6,944 KB
testcase_31 AC 2 ms
6,940 KB
testcase_32 AC 2 ms
6,944 KB
testcase_33 AC 2 ms
6,940 KB
testcase_34 AC 2 ms
6,944 KB
testcase_35 AC 2 ms
6,944 KB
testcase_36 AC 2 ms
6,940 KB
testcase_37 AC 2 ms
6,944 KB
testcase_38 AC 2 ms
6,940 KB
testcase_39 AC 2 ms
6,940 KB
testcase_40 AC 2 ms
6,944 KB
testcase_41 AC 2 ms
6,944 KB
testcase_42 AC 2 ms
6,940 KB
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>
#include<atcoder/all>
#define rep(i,n) for(int i=0;i<n;i++)
using namespace std;
using namespace atcoder;
typedef long long ll;
typedef vector<int> vi;
typedef vector<long long> vl;
typedef vector<vector<int>> vvi;
typedef vector<vector<long long>> vvl;
typedef long double ld;
typedef pair<int, int> P;

template <int m> ostream& operator<<(ostream& os, const static_modint<m>& a) {os << a.val(); return os;}
template <int m> ostream& operator<<(ostream& os, const dynamic_modint<m>& a) {os << a.val(); return os;}
template <int m> istream& operator>>(istream& is, static_modint<m>& a) {long long x; is >> x; a = x; return is;}
template <int m> istream& operator>>(istream& is, dynamic_modint<m>& a) {long long x; is >> x; a = x; return is;}
template<typename T> istream& operator>>(istream& is, vector<T>& v){int n = v.size(); assert(n > 0); rep(i, n) is >> v[i]; return is;}
template<typename U, typename T> ostream& operator<<(ostream& os, const pair<U, T>& p){os << p.first << ' ' << p.second; return os;}
template<typename T> ostream& operator<<(ostream& os, const vector<T>& v){int n = v.size(); rep(i, n) os << v[i] << (i == n - 1 ? "\n" : " "); return os;}
template<typename T> ostream& operator<<(ostream& os, const vector<vector<T>>& v){int n = v.size(); rep(i, n) os << v[i] << (i == n - 1 ? "\n" : ""); return os;}
template<typename T> ostream& operator<<(ostream& os, const set<T>& se){for(T x : se) os << x << " "; os << "\n"; return os;}
template<typename T> ostream& operator<<(ostream& os, const unordered_set<T>& se){for(T x : se) os << x << " "; os << "\n"; return os;}
template<typename S, auto op, auto e> ostream& operator<<(ostream& os, const atcoder::segtree<S, op, e>& seg){int n = seg.max_right(0, [](S){return true;}); rep(i, n) os << seg.get(i) << (i == n - 1 ? "\n" : " "); return os;}
template<typename S, auto op, auto e, typename F, auto mapping, auto composition, auto id> ostream& operator<<(ostream& os, const atcoder::lazy_segtree<S, op, e, F, mapping, composition, id>& seg){int n = seg.max_right(0, [](S){return true;}); rep(i, n) os << seg.get(i) << (i == n - 1 ? "\n" : " "); return os;}

template<typename T> void chmin(T& a, T b){a = min(a, b);}
template<typename T> void chmax(T& a, T b){a = max(a, b);}

using mint = modint998244353;

// combination mod prime
// https://youtu.be/8uowVvQ_-Mo?t=6002
// https://youtu.be/Tgd_zLfRZOQ?t=9928
struct modinv {
  int n; vector<mint> d;
  modinv(): n(2), d({0,1}) {}
  mint operator()(int i) {
    while (n <= i) d.push_back(-d[mint::mod()%n]*(mint::mod()/n)), ++n;
    return d[i];
  }
  mint operator[](int i) const { return d[i];}
} invs;
struct modfact {
  int n; vector<mint> d;
  modfact(): n(2), d({1,1}) {}
  mint operator()(int i) {
    while (n <= i) d.push_back(d.back()*n), ++n;
    return d[i];
  }
  mint operator[](int i) const { return d[i];}
} facts;
struct modfactinv {
  int n; vector<mint> d;
  modfactinv(): n(2), d({1,1}) {}
  mint operator()(int i) {
    while (n <= i) d.push_back(d.back()*invs(n)), ++n;
    return d[i];
  }
  mint operator[](int i) const { return d[i];}
} ifacts;
mint comb(int n, int k) {
  if (n < k || k < 0) return 0;
  return facts(n)*ifacts(k)*ifacts(n-k);
}

mint c = 5;
template<typename mint>
struct Ring{
	mint p, q;
	Ring(mint p = 0, mint q = 0) : p(p), q(q) {}
	bool operator==(const Ring & r) const{return p == r.p && q == r.q;}
	bool operator!=(const Ring & r) const{return p != r.p || q != r.q;}
	Ring operator+(const Ring &r) const { return Ring(*this) += r; }
	Ring operator-(const Ring &r) const { return Ring(*this) -= r; }
	Ring operator*(const Ring &r) const { return Ring(*this) *= r; }
	Ring operator/(const Ring &r) const { return Ring(*this) /= r; }
	Ring &operator+=(const Ring &r) {
		return *this = Ring(p + r.p, q + r.q);
	}
	Ring &operator-=(const Ring &r) {
		return *this = Ring(p - r.p, q - r.q);
	}
	Ring &operator*=(const Ring &r) {
		return *this = Ring(p * r.p + c * q * r.q, p * r.q + q * r.p);
	}
	Ring &operator/=(const Ring &r) {
		*this *= r.inv();
		return *this;
	}
	Ring inv() const{
		Ring res(p / (p * p - c * q * q), -q / (p * p - c * q * q));
		return res;
	}
	Ring pow(long long n) const {
		assert(0 <= n);
		Ring x = *this, r(1, 0);
		while (n) {
			if (n & 1) r *= x;
			x *= x;
			n >>= 1;
		}
		return r;
	}
};

template<typename T> ostream& operator<<(ostream& os, const Ring<T>& r){os << r.p << ':' << r.q; return os;}

using R = Ring<mint>;

int main(){
	long long n;
	int k;
	cin >> n >> k;
	R sum(0, 0);
	mint half = mint(1) / 2;
	for(int i = 0; i <= k; i++){
		R v = R(half, half).pow(i) * R(half, -half).pow(k - i);
		if(v == R(1, 0)){
			sum += R(n, 0) * mint(-1).pow(k - i) * comb(k, i);
		}else{
			sum += (R(1, 0) - v.pow(n)) / (R(1, 0) - v) * v * mint(-1).pow(k - i) * comb(k, i);
		}
	}
	sum /= R(0, 1).pow(k);
	cout << sum.p << "\n";
	return 0;
}
0