結果

問題 No.535 自然数の収納方法
ユーザー antaanta
提出日時 2017-06-23 23:15:12
言語 C++14
(gcc 12.3.0 + boost 1.83.0)
結果
AC  
実行時間 80 ms / 2,000 ms
コード長 4,685 bytes
コンパイル時間 1,713 ms
コンパイル使用メモリ 179,252 KB
実行使用メモリ 19,072 KB
最終ジャッジ日時 2024-10-03 03:35:48
合計ジャッジ時間 3,335 ms
ジャッジサーバーID
（参考情報） 		judge5 / judge3
このコードへのチャレンジ
（要ログイン）

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
6,820 KB
testcase_01 AC 2 ms
6,820 KB
testcase_02 AC 2 ms
6,820 KB
testcase_03 AC 2 ms
6,816 KB
testcase_04 AC 2 ms
6,820 KB
testcase_05 AC 2 ms
6,816 KB
testcase_06 AC 2 ms
6,820 KB
testcase_07 AC 14 ms
6,816 KB
testcase_08 AC 46 ms
12,416 KB
testcase_09 AC 60 ms
14,712 KB
testcase_10 AC 70 ms
17,024 KB
testcase_11 AC 9 ms
6,820 KB
testcase_12 AC 27 ms
8,704 KB
testcase_13 AC 2 ms
6,820 KB
testcase_14 AC 20 ms
7,296 KB
testcase_15 AC 71 ms
17,792 KB
testcase_16 AC 43 ms
12,288 KB
testcase_17 AC 77 ms
18,432 KB
testcase_18 AC 2 ms
6,816 KB
testcase_19 AC 80 ms
18,944 KB
testcase_20 AC 79 ms
18,944 KB
testcase_21 AC 76 ms
19,072 KB
testcase_22 AC 79 ms
19,072 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"
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<1000000007> mint;

#pragma region for precomputing
int berlekampMassey(const vector<mint> &s, vector<mint> &C) {
	int N = (int)s.size();
	C.assign(N + 1, mint());
	vector<mint> B(N + 1, mint());
	C[0] = B[0] = 1;
	int degB = 0;
	vector<mint> T;
	int L = 0, m = 1;
	mint b = 1;
	for (int n = 0; n < N; ++ n) {
		mint d = s[n];
		for (int i = 1; i <= L; ++ i)
			d += C[i] * s[n - i];
		if (d == mint()) {
			++ m;
		} else {
			if (2 * L <= n)
				T.assign(C.begin(), C.begin() + (L + 1));
			mint coeff = -d * b.inverse();
			for (int i = 0; i <= degB; ++ i)
				C[m + i] += coeff * B[i];
			if (2 * L <= n) {
				L = n + 1 - L;
				B.swap(T);
				degB = (int)B.size() - 1;
				b = d;
				m = 1;
			} else {
				++ m;
			}
		}
	}
	C.resize(L + 1);
	return L;
}

void computeMinimumPolynomialForLinearlyRecurrentSequence(const vector<mint> &a, vector<mint> &phi) {
	int n2 = (int)a.size(), n = n2 / 2;
	assert(n2 % 2 == 0);
	int L = berlekampMassey(a, phi);
	reverse(phi.begin(), phi.begin() + (L + 1));
}

template<int MOD> int mintToSigned(ModInt<MOD> a) {
	int x = a.get();
	if (x <= MOD / 2)
		return x;
	else
		return x - MOD;
}

int outputPrecomputedMinimalPolynomial(vector<mint> seq, ostream &os) {
	if (seq.size() % 2 == 1) seq.pop_back();
	vector<mint> phi;
	computeMinimumPolynomialForLinearlyRecurrentSequence(seq, phi);
	if (phi.size() >= seq.size() / 2 - 2) {
		cerr << "warning: maybe not enough terms" << endl;
	}
	cerr << "/*" << phi.size() - 1 << "*/";
	os << "{{ ";
	rep(i, phi.size() - 1) {
		if (i != 0) os << ", ";
		os << seq[i].get();
	}
	os << " }, { ";
	rep(i, phi.size()) {
		if (i != 0) os << ", ";
		os << mintToSigned(phi[i]);
	}
	os << " }}";
	return (int)phi.size() - 1;
}

#pragma endregion

int main() {
	//A_{i-1} - (i - 1) < A_i
	//A_{i-1} - i < A_i
	//A_N - N < A_1
	//A_N <= A_1
	//A_{N-1} - {N-1} < A_N
	//A_{N-1} < A_N + N
	int N;
	while (~scanf("%d", &N)) {
		mint ans;
		rep(ANis1, 2) {
			vector<vector<mint>> dp(N - 1, vector<mint>(N + 2));
			if (ANis1) {
				dp[0][1] += 1;
			} else {
				rer(AN, 2, N)
					dp[0][AN] += 1;
			}
			rep(i, N - 2) {
				rer(j, 1, N) {
					mint x = dp[i][j];
					if (x.x == 0) continue;
					if (i + 1 <= j) {
						dp[i + 1][j - i] += x;
						dp[i + 1][N + 1 - i] -= x;
					} else {
						//k = 1 + i
						dp[i + 1][1] += x * (i + 1 - j);
						dp[i + 1][2] -= x * (i + 1 - j);
						dp[i + 1][1] += x;
						dp[i + 1][N + 1 - i] -= x;
					}
				}
				rer(j, 1, N)
					dp[i + 1][j + 1] += dp[i + 1][j];
			}
			mint sum;
			rer(j, 1, N) {
				mint x = dp[N - 2][j];
				if (x.x == 0) continue;
				sum += x * (N - j);
				if (!ANis1)
					sum += x;
			}
			ans += sum;
		}
		printf("%d\n", ans.get());
	}
	return 0;
}
0