Matrix Fast Power

Reference: http://www.cnblogs.com/yan-boy/archive/2012/11/29/2795294.html

For a^n, we generally pass the n-1, but we can reduce the number of times by making simple improvements using the combination of matrix multiplication, for example, A*a*a*a*a*a = (a*a) * (a*a) * (a*a), which reduces the number of times from 5 to 3. So how do you get the minimum number of times to use the binding law?

The answer is: the combination of binary and dichotomy ideas

For example, a^21 = (a^16) * (a^4) * (a^1) (21=2^4+2^2+2^0), a^16 can be divided into two (a^8) * (a^8), a^8= (a^4) * (a^4) ....

Thus the number of factors can be reduced from N to log (n)

#include <stdio.h>#include<stdlib.h>#include<string.h>#include<time.h>#defineN 100#definell Long Longstructmat{ll mat[n][n];};voidShowmat (Mat a) { for(intI=0;i<3; i++)    {         for(intj=0;j<3; j + +) printf ("%lld", A.mat[i][j]); printf ("\ n"); }}voidInit (Mat &b)//initialized to the unit matrix, to reference{    intI, j, K;  for(i =0; I <3; ++i) { for(j =0; J <3; ++j) {B.mat[i][j]= (i = = j);//a shorthand notation}}}mat Matmul (Mat A,mat b) {mat C; memset (C.mat,0,sizeof(C.mat)); intI, j, K;  for(i =0; I <3; ++i) { for(j =0; J <3; ++j) { for(k =0; K <3; ++k) {C.mat[i][j]+ = a.mat[i][k] *B.mat[k][j]; }        }    }    returnC;} Mat MATLAB (Mat A,intN//Core Code, example a^156= (a^4) * (a^8) * (a^16) * (a^128){Mat B;    Init (b);  while(n) {if(n&1) b=Matmul (b,a); N>>=1;//Move RightA =Matmul (a,a); }    returnb;}intMain () {intN;    Mat A, B; scanf ("%d",&N); memset (A.mat,0,sizeof(A.mat)); memset (B.mat,0,sizeof(B.mat)); //generating three-dimensional random matricesSrand (Time (0));//Random seed---stdlib.h     for(intI=0;i<3; i++)         for(intj=0;j<3; j + +) A.mat[i][j]= rand ()% -; printf ("the random matrix is: \ n");    Showmat (a); b=MATLAB (a,n); printf ("the resulting matrix of%d powers is: \ n", N);    Showmat (b); return 0;}

Matrix Fast Power