結果

問題 No.3243 Multiplication 8 1
ユーザー 遭難者
提出日時 2025-07-07 14:15:57
言語 C++17
(gcc 13.3.0 + boost 1.87.0)
結果
AC  
実行時間 86 ms / 2,000 ms
コード長 4,721 bytes
コンパイル時間 4,499 ms
コンパイル使用メモリ 257,760 KB
実行使用メモリ 7,844 KB
最終ジャッジ日時 2025-07-10 22:22:49
合計ジャッジ時間 4,510 ms
ジャッジサーバーID
(参考情報)
judge1 / judge3
このコードへのチャレンジ
(要ログイン)
ファイルパターン 結果
sample AC * 1
other AC * 4
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 std::cerr;
using std::cin;
using std::cout;
#if __has_include(<atcoder/all>)
#include <atcoder/all>
using mint = atcoder::modint998244353;
istream &operator>>(istream &is, mint &a) {
	int t;
	is >> t;
	a = t;
	return is;
}
ostream &operator<<(ostream &os, mint a) { return os << a.val(); }
#endif
typedef long double ld;
#define long long long
#define uint unsigned int
#define ull unsigned long
#define overload3(a, b, c, name, ...) name
#define rep3(i, a, b) for (int i = (a); i < (b); i++)
#define rep2(i, n) rep3(i, 0, n)
#define rep1(n) rep2(i, n)
#define rep(...) overload3(__VA_ARGS__, rep3, rep2, rep1)(__VA_ARGS__)
#define per3(i, a, b) for (int i = (b) - 1; i >= (a); i--)
#define per2(i, n) per3(i, 0, n)
#define per1(n) per2(i, n)
#define per(...) overload3(__VA_ARGS__, per3, per2, per1)(__VA_ARGS__)
#define all(a) a.begin(), a.end()
#define UNIQUE(a)                                                              \
	sort(all(a));                                                              \
	a.erase(unique(all(a)), a.end())
#define sz(a) (int)a.size()
#define vec vector
#ifndef DEBUG
#define cerr                                                                   \
	if (0)                                                                     \
	cerr
// #undef assert
// #define assert(...) void(0)
#undef endl
#define endl '\n'
#endif
template <typename T> ostream &operator<<(ostream &os, vector<T> a) {
	const int n = a.size();
	rep(i, n) {
		os << a[i];
		if (i + 1 != n)
			os << " ";
	}
	return os;
}
template <typename T, size_t n>
ostream &operator<<(ostream &os, array<T, n> a) {
	rep(i, n) {
		os << a[i];
		if (i + 1 != n)
			os << " ";
	}
	return os;
}
template <typename T> istream &operator>>(istream &is, vector<T> &a) {
	for (T &i : a)
		is >> i;
	return is;
}
template <typename T, typename S> bool chmin(T &x, S y) {
	if ((T)y < x) {
		x = (T)y;
		return true;
	}
	return false;
}
template <typename T, typename S> bool chmax(T &x, S y) {
	if (x < (T)y) {
		x = (T)y;
		return true;
	}
	return false;
}
template <typename T> void operator++(vector<T> &a) {
	for (T &i : a)
		++i;
}
template <typename T> void operator--(vector<T> &a) {
	for (T &i : a)
		--i;
}
template <typename T> void operator++(vector<T> &a, int) {
	for (T &i : a)
		i++;
}
template <typename T> void operator--(vector<T> &a, int) {
	for (T &i : a)
		i--;
}

template <typename T> struct M {
	vector<vector<T>> a;
	int n, m;
	M(int n, int m) : n(n), m(m), a(n, vector<T>(m)) {}
	M(int n = 0) : M<T>(n, n) {}
	vector<T> &operator[](int k) { return a[k]; }
	const vector<T> &operator[](int k) const { return a[k]; }
	static M I(int n) {
		M mat(n);
		rep(i, n) mat[i][i] = 1;
		return mat;
	}
	M &operator+=(const M &b) {
		rep(i, n) rep(j, m)(*this)[i][j] += b[i][j];
		return *this;
	}
	M &operator-=(const M &b) {
		rep(i, n) rep(j, m)(*this)[i][j] -= b[i][j];
		return *this;
	}
	M &operator*=(const M &b) {
		int l = b.m;
		vector c(n, vector<T>(l));
		rep(i, n) rep(j, m) rep(k, l) c[i][k] += (*this)[i][j] * b[j][k];
		a.swap(c);
		return *this;
	}
	M &operator^=(long k) {
		M b = M::I(n);
		while (k) {
			if (k & 1)
				b *= *this;
			*this *= *this;
			k >>= 1;
		}
		a.swap(b.a);
		return *this;
	}
	M operator+(const M &b) const { return (M(*this) += b); }
	M operator-(const M &b) const { return (M(*this) -= b); }
	M operator*(const M &b) const { return (M(*this) *= b); }
	M operator^(const M &b) const { return (M(*this) ^= b); }
};
template <typename T> pair<int, T> GaussElimination(M<T> &a, bool LE = false) {
	int n = a.n, m = a.m;
	int rank = 0, je = LE ? m - 1 : m;
	mint det = 1;
	rep(j, je) {
		int idx = -1;
		rep(i, rank, n) {
			if (a[i][j].x) {
				idx = i;
				break;
			}
		}
		if (idx == -1) {
			det = 0;
			continue;
		}
		if (rank != idx) {
			det = -det;
			swap(a[rank], a[idx]);
		}
		det *= a[rank][j];
		if (LE && a[rank][j].x != 1) {
			mint coeff = a[rank][j].inv();
			rep(k, j, m) a[rank][k] *= coeff;
		}
		int is = LE ? 0 : rank + 1;
		rep(i, is, n) {
			if (i == rank)
				continue;
			if (a[i][j].x) {
				mint coeff = a[i][j] / a[rank][j];
				rep(k, j, m) a[i][k] -= a[rank][k] * coeff;
			}
		}
		rank++;
	}
	return make_pair(rank, det);
}

void solve() {
	M<mint> a(7);
	rep(i, 6) {
		a[i][i] = 1;
		a[i ^ 1][i] = 1;
		if (i < 4) {
			a[i + 2][i] = 1;
			a[(i + 2) ^ 1][i] = 1;
		}
	}
	a[6][4] = a[6][5] = 1;
	a[0][4] = a[0][5] = 1;
	a[6][6] = a[0][6] = 1;
	long n;
	cin >> n;
	a ^= n;
	cout << a[0][0] - mint(2).pow(n - 1) << endl;
}
int main() {
	// srand((unsigned)time(NULL));
	cin.tie(nullptr);
	ios::sync_with_stdio(false);
	cout << fixed << setprecision(20);
	int t = 1;
	cin >> t;
	while (t--)
		solve();
}
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