Rapid power and Application of Matrices

Take the second-order matrix as an example. A typical Fibonacci series can be solved by multiplication of matrices.

If multiplication is not considered, the complexity should be lgn.

The following describes how to calculate the power of a matrix.

# Include <iostream> # define l 10000 using namespace STD; Class mymatrix {public: long a00, A01, A10, A11; mymatrix (); mymatrix (long _ a00, long long _ A01, long _ A10, long _ A11): a00 (_ a00), A01 (_ A01), A10 (_ A10), A11 (_ A11) {} void print_matrix () ;}; void mymatrix: print_matrix () {cout <a00 <"" <A01 <Endl <A10 <"" <A11 <Endl;} const mymatrix operator * (const mymatrix & M1, const mymatrix & m2) {return mymatrix (m1.a00 * m2.a00 + m1.a01 * m2.a10, m1.a00 * m2.a01 + signature * m2.a11, m1.a10 * m2.a00 + m1.a11 * m2.a10, m1.a10 * m2.a01 + m1.a11 * m2.a11);} // implement Recursive Sub-governance to solve const mymatrix pow_matrix (const mymatrix & M, long n) {If (n <= 1) return m; else {If (N & 1) {// odd mymatrix temp = pow_matrix (m, n/2); temp = temp * m; return temp ;} else {mymatrix temp = pow_matrix (m, n/2); temp = temp * temp; return temp ;}}// practically recursive optimization solution mymatrix ans2, 0, 1); // defines a unit array. The final result void pow_matrix_fast (const mymatrix & M, long n) {mymatrix temp = m; // temp. print_matrix (); While (n) {If (N & 1) {// odd number of ans2 = ans2 * temp;} temp = temp * temp; n >>= 1; // temp. print_matrix () ;}} mymatrix ans3 (,); void pow_matrix_fast2 (const mymatrix & M, long n) {mymatrix temp = m; // temp. print_matrix (); n --; while (n) {If (N & 1) {// odd ans3 = ans3 * temp;} temp = temp * temp; n> = 1; // temp. print_matrix () ;}} int main () {mymatrix M (1, 2, 3, 4); mymatrix ans = pow_matrix (M, 6); ans. print_matrix (); pow_matrix_fast (M, 6); ans2.print _ matrix (); pow_matrix_fast2 (M, 6); ans3.print _ matrix (); Return 0 ;}