【矩阵快速幂】hdu Another kind of Fibonacci

【矩阵快速幂】hdu  Another kind of Fibonacci

#include<iostream>
using namespace std;
typedef long long ll;
const ll mod = 10007;
ll N, X, Y;

struct Matrix {
	static const int N = 15;
	ll a[N][N];
	Matrix(ll e = 0) {
		for (int i = 1; i <= 4; i++)
			for (int j = 1; j <= 4; j++)
				a[i][j] = e * (i == j);
	}
	Matrix mul(Matrix A, Matrix B) {
		Matrix ans(0);
		for (int i = 1; i <= 4; i++) {
			for (int j = 1; j <= 4; j++) {
				for (int k = 1; k <= 4; k++) {
					ans.a[i][j] = (ans.a[i][j] + A.a[i][k] * B.a[k][j]) % mod;
				}
			}
		}
		return ans;
	}
	Matrix ksm(Matrix A, ll b) {
		Matrix ans(1);
		while (b) {
			if (b & 1)ans = mul(ans, A);
			A = mul(A, A); b >>= 1;
		}
		return ans;
	}
};

ll c[5];
int main() {
	ios::sync_with_stdio(false);
	cin.tie(0); cout.tie(0);

	c[1] = 2; c[2] = 1; c[3] = 1; c[4] = 1;
	while (cin >> N >> X >> Y) {
		Matrix tmp;
		if (N < 2) {
			cout << N + 1 << "\n";
			continue;
		}
		tmp.a[1][1] = 1;
		tmp.a[1][2] = X * X % mod;
		tmp.a[1][3] = 2 * X * Y % mod;
		tmp.a[1][4] = Y * Y % mod;
		tmp.a[2][2] = X * X % mod;
		tmp.a[2][3] = 2 * X * Y % mod;
		tmp.a[2][4] = Y * Y % mod;
		tmp.a[3][2] = X % mod;
		tmp.a[3][3] = Y % mod;
		tmp.a[4][2] = 1;
		tmp = tmp.ksm(tmp, N - 1);
		ll res = 0;
		for (int i = 1; i <= 4; ++i)
			res = (res + tmp.a[1][i] * c[i]) % mod;
		cout << res << "\n";
	}
}

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