一个简单的矩阵乘法（快速幂）

　　//求FIB的第n项
 1 #include <iostream>
 2 #include <vector>
 3 #include <cstdio>
 4 
 5 using namespace std;
 6 #define LL long long
 7 
 8 typedef vector<int> vec;
 9 typedef vector<vec> mat;
10 
11 mat mul(mat& A, mat& B)
12 {
13     mat C ( A.size(),vec(B[0].size()));
14 
15     for(int i=0;i<A.size();i++)
16     {
17         for(int k = 0;k<B.size();k++)
18         {
19             for(int j = 0;j<B[0].size();j++)
20             {
21                 C[i][j] = C[i][j] + A[i][k]*B[k][j] ;
22             }
23         }
24     }
25     return C;
26 }
27 
28 mat pow(mat A, LL n)
29 {
30     mat B( A.size(),vec(A.size()));
31 
32     for(int i=0;i<A.size();i++)
33     {
34         B[i][i] =1;
35     }
36     while( n>0)
37     {
38         if( n&1) B = mul(B,A);
39         A = mul(A,A);
40         n>>=1;
41     }
42     return B;
43 
44 
45 }
46 
47 int main()
48 {
49     LL n ;
50     cin>>n;
51     mat A ( 2, vec(2));
52     A[0][0] =1; A[0][1] = 1;
53     A[1][0] = 1; A[1][1] = 0;
54     A = pow( A, n);
55     printf("%d\n",A[1][0]);
56     return 0;
57 }

 

转载于:https://www.cnblogs.com/doubleshik/p/3537666.html