結果

問題 No.658 テトラナッチ数列 Hard
ユーザー kissshot7kissshot7
提出日時 2020-08-17 15:24:14
言語 C++14
(gcc 12.3.0 + boost 1.83.0)
結果
AC  
実行時間 616 ms / 2,000 ms
コード長 3,690 bytes
コンパイル時間 1,906 ms
コンパイル使用メモリ 178,480 KB
実行使用メモリ 7,552 KB
最終ジャッジ日時 2024-10-11 11:04:32
合計ジャッジ時間 5,207 ms
ジャッジサーバーID
（参考情報） 		judge1 / judge4
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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 4 ms
7,552 KB
testcase_01 AC 5 ms
7,424 KB
testcase_02 AC 4 ms
7,492 KB
testcase_03 AC 6 ms
7,492 KB
testcase_04 AC 224 ms
7,492 KB
testcase_05 AC 257 ms
7,424 KB
testcase_06 AC 324 ms
7,424 KB
testcase_07 AC 350 ms
7,424 KB
testcase_08 AC 407 ms
7,424 KB
testcase_09 AC 611 ms
7,488 KB
testcase_10 AC 616 ms
7,552 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 int long long
typedef long long ll;

typedef unsigned long long ul;
typedef unsigned int ui;
const ll mod = 1000000007;
const ll INF = mod * mod;
const int INF_N = 1e+9;
typedef pair<int, int> P;
#define stop char nyaa;cin>>nyaa;
#define rep(i,n) for(int i=0;i<n;i++)
#define per(i,n) for(int i=n-1;i>=0;i--)
#define Rep(i,sta,n) for(int i=sta;i<n;i++)
#define rep1(i,n) for(int i=1;i<=n;i++)
#define per1(i,n) for(int i=n;i>=1;i--)
#define Rep1(i,sta,n) for(int i=sta;i<=n;i++)
#define all(v) (v).begin(),(v).end()
typedef pair<ll, ll> LP;
typedef long double ld;
typedef pair<ld, ld> LDP;
const ld eps = 1e-12;
const ld pi = acos(-1.0);
//typedef vector<vector<ll>> mat;
typedef vector<int> vec;

//繰り返し二乗法
ll mod_pow(ll a, ll n, ll m) {
	ll res = 1;
	while (n) {
		if (n & 1)res = res * a%m;
		a = a * a%m; n >>= 1;
	}
	return res;
}

struct modint {
	ll n;
	modint() :n(0) { ; }
	modint(ll m) :n(m) {
		if (n >= mod)n %= mod;
		else if (n < 0)n = (n%mod + mod) % mod;
	}
	operator int() { return n; }
};
bool operator==(modint a, modint b) { return a.n == b.n; }
modint operator+=(modint &a, modint b) { a.n += b.n; if (a.n >= mod)a.n -= mod; return a; }
modint operator-=(modint &a, modint b) { a.n -= b.n; if (a.n < 0)a.n += mod; return a; }
modint operator*=(modint &a, modint b) { a.n = ((ll)a.n*b.n) % mod; return a; }
modint operator+(modint a, modint b) { return a += b; }
modint operator-(modint a, modint b) { return a -= b; }
modint operator*(modint a, modint b) { return a *= b; }
modint operator^(modint a, int n) {
	if (n == 0)return modint(1);
	modint res = (a*a) ^ (n / 2);
	if (n % 2)res = res * a;
	return res;
}

//逆元(Eucledean algorithm)
ll inv(ll a, ll p) {
	return (a == 1 ? 1 : (1 - p * inv(p%a, a)) / a + p);
}
modint operator/(modint a, modint b) { return a * modint(inv(b, mod)); }

const int max_n = 1 << 18;
modint fact[max_n], factinv[max_n];
void init_f() {
	fact[0] = modint(1);
	for (int i = 0; i < max_n - 1; i++) {
		fact[i + 1] = fact[i] * modint(i + 1);
	}
	factinv[max_n - 1] = modint(1) / fact[max_n - 1];
	for (int i = max_n - 2; i >= 0; i--) {
		factinv[i] = factinv[i + 1] * modint(i + 1);
	}
}
modint comb(int a, int b) {
	if (a < 0 || b < 0 || a < b)return 0;
	return fact[a] * factinv[b] * factinv[a - b];
}
using mP = pair<modint, modint>;

int dx[4] = { 0,1,0,-1 };
int dy[4] = { 1,0,-1,0 };


struct Matrix {
	vector<vector<long long> > val;
	Matrix(int n, int m, long long x = 0) : val(n, vector<long long>(m, x)) {}
	void init(int n, int m, long long x = 0) { val.assign(n, vector<long long>(m, x)); }
	size_t size() const { return val.size(); }
	inline vector<long long>& operator [] (int i) { return val[i]; }
};

Matrix operator * (Matrix A, Matrix B) {
	Matrix R(A.size(), B[0].size());
	for (int i = 0; i < A.size(); ++i)
		for (int j = 0; j < B[0].size(); ++j)
			for (int k = 0; k < B.size(); ++k)
				R[i][j] = (R[i][j] + A[i][k] * B[k][j] % 17) % 17;
	return R;
}

Matrix modpow(Matrix A, long long n) {
	Matrix R(A.size(), A.size());
	for (int i = 0; i < A.size(); ++i) R[i][i] = 1;
	while (n > 0) {
		if (n & 1) R = R * A;
		A = A * A;
		n >>= 1;
	}
	return R;
}

void solve() {
    int q; cin >> q;
    Matrix A(4, 4);
    // A[0][0] = A[0][1] = A[0][2] = A[0][3] = A[1][0] = A[2][1] = A[3][2] = 1;
    A.val = {{1,1,1,1}, {1,0,0,0}, {0,1,0,0}, {0,0,1,0}};
    rep(i, q){
        ll n; cin >> n;
        cout << modpow(A, n)[3][3] << endl;
    }
}

signed main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  //cout << fixed << setprecision(10);
  //init_f();
  //init();
  //int t; cin >> t; rep(i, t)solve();
  solve();
//   stop
    return 0;
}
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