[矩阵快速幂] 数列（类斐波那契

 1 #include<cstdio>
 2 #include<cstring>
 3  
 4 const int Mod = 1e9 + 7; 
 5  
 6 struct Matrix {
 7     long long m[3][3]; 
 8 }; 
 9  
10 Matrix Mult(Matrix a, Matrix b) {
11     long long sums = 0; 
12     Matrix c; 
13     memset(c.m, 0, sizeof(c.m)); 
14     for (int i = 0; i <= 2; i++) {
15         for (int j = 0; j <= 2; j++) {
16             sums = 0; 
17             for (int k = 0; k <= 2; k++) {
18                 sums = (sums + a.m[i][k] * b.m[k][j]) % Mod; 
19             }
20             c.m[i][j] = sums; 
21         }
22     }
23     return c; 
24 }
25  
26 Matrix Qpow(Matrix a, int k) {
27     Matrix res;
28     memset(res.m, 0, sizeof(res.m));
29     for (int i = 0; i <= 2; i++) {
30         res.m[i][i] = 1; 
31     } 
32     while(k) {
33         if (k & 1) {
34             res = Mult(res, a); 
35         }
36         a = Mult(a, a); 
37         k = (k >> 1); 
38     }
39     return res; 
40 }
41  
42 int main() {
43     long long n; 
44     int t; 
45     scanf("%d", &t); 
46     for (int cnt = 1; cnt <= t; cnt++) {
47         Matrix A, B; 
48         memset(A.m, 0, sizeof(A.m)); 
49         memset(B.m, 0, sizeof(B.m)); 
50         scanf("%lld", &n); 
51         if (n <= 3) {
52             printf("1\n"); 
53             continue; 
54         }
55         for (int i = 0; i < 3; i++) {
56             A.m[i][i + 1] = 1; 
57         }
58         A.m[2][0] = 1;
59         A.m[2][2] = 1; 
60         for (int i = 0; i < 3; i++) {
61             B.m[i][0] = 1; 
62         }
63         A = Qpow(A, n - 3); 
64         A = Mult(A, B); 
65         printf("%d\n", A.m[2][0]); 
66     }
67     return 0; 
68 }