結果

問題 No.535 自然数の収納方法
ユーザー anta
提出日時 2017-06-23 23:15:12
言語 C++14
(gcc 13.3.0 + boost 1.87.0)
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
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ファイルパターン 結果
other AC * 23
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ソースコード

diff #
プレゼンテーションモードにする

#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;
}
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