矩阵快速幂

 1 typedef vector<int> vec;
 2 typedef vector<vec> mat;
 3 
 4 //计算A*B
 5 mat mul(mat &A, mat &B) {
 6     mat C(A.size(), vec(B[0].size()));
 7     for (int i = 0; i < A.size(); i++) {
 8         for (int k = 0; k < B.size(); k++) {
 9             for (int j = 0; j < B[0].size(); j++) {
10                 C[i][j] = (C[i][j] + A[i][k] * B[k][j]) % MOD;
11             }
12         }
13     }
14     return C;
15 }
16 
17 //计算A^n
18 mat mat_pow(mat A, LL n) {
19     mat B(A.size(), vec(A.size()));
20     for (int i = 0; i < A.size(); i++) {
21         B[i][i] = 1;
22     }
23     while (n > 0) {
24         if (n & 1) B = mul(B, A);
25         A = mul(A, A);
26         n >>= 1;
27     }
28     return B;
29 }