結果
| 問題 |
No.658 テトラナッチ数列 Hard
|
| コンテスト | |
| ユーザー |
anta
|
| 提出日時 | 2018-03-02 22:27:22 |
| 言語 | C++14 (gcc 13.3.0 + boost 1.87.0) |
| 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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| ファイルパターン | 結果 |
|---|---|
| sample | AC * 3 |
| other | AC * 8 |
ソースコード
#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;
}
anta