結果

問題 No.658 テトラナッチ数列 Hard
ユーザー antaanta
提出日時 2018-03-02 22:27:22
言語 C++14
(gcc 12.3.0 + boost 1.83.0)
結果
AC  
実行時間 79 ms / 2,000 ms
コード長 3,519 bytes
コンパイル時間 2,178 ms
コンパイル使用メモリ 174,732 KB
実行使用メモリ 5,376 KB
最終ジャッジ日時 2024-06-13 03:27:36
合計ジャッジ時間 2,796 ms
ジャッジサーバーID
（参考情報） 		judge5 / 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
5,248 KB
testcase_01 AC 2 ms
5,376 KB
testcase_02 AC 2 ms
5,376 KB
testcase_03 AC 2 ms
5,376 KB
testcase_04 AC 28 ms
5,376 KB
testcase_05 AC 33 ms
5,376 KB
testcase_06 AC 41 ms
5,376 KB
testcase_07 AC 44 ms
5,376 KB
testcase_08 AC 51 ms
5,376 KB
testcase_09 AC 79 ms
5,376 KB
testcase_10 AC 77 ms
5,376 KB
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ソースコード

diff # 

#include "bits/stdc++.h"
using namespace std;
#define rep(i,n) for(int (i)=0;(i)<(int)(n);++(i))
#define rer(i,l,u) for(int (i)=(int)(l);(i)<=(int)(u);++(i))
#define reu(i,l,u) for(int (i)=(int)(l);(i)<(int)(u);++(i))
static const int INF = 0x3f3f3f3f; static const long long INFL = 0x3f3f3f3f3f3f3f3fLL;
typedef vector<int> vi; typedef pair<int, int> pii; typedef vector<pair<int, int> > vpii; typedef long long ll;
template<typename T, typename U> static void amin(T &x, U y) { if (y < x) x = y; }
template<typename T, typename U> static void amax(T &x, U y) { if (x < y) x = y; }


template<int MOD>
struct ModInt {
	static const int Mod = MOD;
	unsigned x;
	ModInt() : x(0) { }
	ModInt(signed sig) { int sigt = sig % MOD; if (sigt < 0) sigt += MOD; x = sigt; }
	ModInt(signed long long sig) { int sigt = sig % MOD; if (sigt < 0) sigt += MOD; x = sigt; }
	int get() const { return (int)x; }

	ModInt &operator+=(ModInt that) { if ((x += that.x) >= MOD) x -= MOD; return *this; }
	ModInt &operator-=(ModInt that) { if ((x += MOD - that.x) >= MOD) x -= MOD; return *this; }
	ModInt &operator*=(ModInt that) { x = (unsigned long long)x * that.x % MOD; return *this; }
	ModInt &operator/=(ModInt that) { return *this *= that.inverse(); }

	ModInt operator+(ModInt that) const { return ModInt(*this) += that; }
	ModInt operator-(ModInt that) const { return ModInt(*this) -= that; }
	ModInt operator*(ModInt that) const { return ModInt(*this) *= that; }
	ModInt operator/(ModInt that) const { return ModInt(*this) /= that; }

	ModInt inverse() const {
		signed a = x, b = MOD, u = 1, v = 0;
		while (b) {
			signed t = a / b;
			a -= t * b; std::swap(a, b);
			u -= t * v; std::swap(u, v);
		}
		if (u < 0) u += Mod;
		ModInt res; res.x = (unsigned)u;
		return res;
	}

	bool operator==(ModInt that) const { return x == that.x; }
	bool operator!=(ModInt that) const { return x != that.x; }
	ModInt operator-() const { ModInt t; t.x = x == 0 ? 0 : Mod - x; return t; }
};
typedef ModInt<17> mint;

mint linearlyRecurrentSequenceValue(long long K, const vector<mint> &initValues, const vector<mint> &annPoly) {
	assert(K >= 0);
	if (K < (int)initValues.size())
		return initValues[(int)K];
	int d = (int)annPoly.size() - 1;
	assert(d >= 0);
	assert(annPoly[d].get() == 1);
	assert(d <= (int)initValues.size());
	if (d == 0)
		return mint();
	vector<mint> coeffs(d), square;
	coeffs[0] = 1;
	int l = 0;
	while ((K >> l) > 1) ++ l;
	for (; l >= 0; -- l) {
		square.assign(d * 2 - 1, mint());
		for (int i = 0; i < d; ++ i)
			for (int j = 0; j < d; ++ j)
				square[i + j] += coeffs[i] * coeffs[j];
		for (int i = d * 2 - 2; i >= d; -- i) {
			mint c = square[i];
			if (c.x == 0) continue;
			for (int j = 0; j < d; ++ j)
				square[i - d + j] -= c * annPoly[j];
		}
		for (int i = 0; i < d; ++ i)
			coeffs[i] = square[i];
		if (K >> l & 1) {
			mint lc = coeffs[d - 1];
			for (int i = d - 1; i >= 1; -- i)
				coeffs[i] = coeffs[i - 1] - lc * annPoly[i];
			coeffs[0] = mint() - lc * annPoly[0];
		}
	}
	mint res;
	for (int i = 0; i < d; ++ i)
		res += coeffs[i] * initValues[i];
	return res;
}

mint linearlyRecurrentSequenceValue(long long K, const pair<vector<mint>, vector<mint> > &seqPair) {
	return linearlyRecurrentSequenceValue(K, seqPair.first, seqPair.second);
}

int main() {
	int Q;
	while (~scanf("%d", &Q)) {
		rep(i, Q) {
			long long n;
			scanf("%lld", &n);
			mint ans = linearlyRecurrentSequenceValue(n, { { 0, 0, 0, 0, 1 },{ 0, -1, -1, -1, -1, 1 } });
			printf("%d\n", ans.get());
		}
	}
	return 0;
}
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