HDOJ 6030 矩阵快速幂

31 typedef vector<ll> vec;
32 typedef vector<vec> mat;
33 
34 mat mul(mat &A, mat &B) {
35     mat C(A.size(), vec(B[0].size()));
36     rep(i, 0, A.size()) rep(k, 0, B.size()) rep(j, 0, B[0].size())
37         C[i][j] = (C[i][j] + A[i][k] * B[k][j]) % MOD;
38     return C;
39 }
40 
41 mat pow(mat A, ll n) {
42     mat B(A.size(), vec(A.size()));
43     rep(i, 0, A.size()) B[i][i] = 1;
44     while (n>0) {
45         if (n & 1) B = mul(B, A);
46         A = mul(A, A);
47         n >>= 1;
48     }
49     return B;
50 }
51 
52 int main() {
53     int T;
54     cin >> T;
55     while (T--) {
56         mat A(3, vec(3));
57         A[0][0] = 1, A[0][1] = 0, A[0][2] = 1;
58         A[1][0] = 1, A[1][1] = 0, A[2][2] = 0;
59         A[2][0] = 0, A[2][1] = 1, A[2][2] = 0;
60         ll n;
61         cin >> n;
62         A = pow(A, n - 2);
63         ll ans = (A[2][0] * 6 + A[2][1] * 4 + A[2][2] * 3) % MOD;
64         cout << ans << endl;
65     }
66     return 0;
67 }