Blue Bridge: Matrix Multiplication

Matrix Multiplication

Problem Description

　Given ⼀ N- order matrix A , the output A of the M -th power ( M is an integer comes in handy negative)

　E.g:

　　A = 1 2

　　       3 4

　　A is a power of 2   710

　　                15 22

   START input format

     ⾏ ⼀ first frame is positive integer N , M ( . 1 <= N <= 30, 0 <= M <=. 5), the matrix represents A a power of order and claimed

　 Next N ⾏, each ⾏ N absolute values does not exceed 10 to comes in handy negative integer, described matrix A value

Output Format

　　The output common N ⾏, each ⾏ N integers, indicates A of M -th power corresponding to the matrix. Using several adjacent spaces separated ⼀

Sample lose START

2 2

1 2

3 4

Sample Output

7 10

15 22

Analysis: There are actually a ⼀! 0 power! Ming matrix to zero output matrix!

#include <iostream>
using namespace std;
long long int b[40][40];
int main() {
 int n, m;
 cin >> n >> m;
 long long int a[40][40];
 long long int t[40][40];
 for(int i = 0; i < n; i++) {
     for(int j = 0; j < n; j++) {
         cin >> a[i][j];
         t[i][j] = a[i][j];
     }
 }
 if( m == 0) {
     for(int i = 0; i < n; i++) {
         for(int j = 0; j < n; j++) {
             if(i != j) {
                cout << 0 << " ";
             } else {
                cout << 1 << " ";
             }
         }  
      cout << endl;
     }
    return 0;
 }
 while(--m) {
    for(int i = 0; i < n; i++) {
        for(int j = 0; j < n; j++) {
            int k = n;
            while(k) {
              b[i][j] += t[i][k-1] * a[k-1][j];
              k--;
            }
        }
    }
    for( int  i = 0; i < n; i++) {
         for( int j = 0; j < n; j++) {
              t[i][j] = b[i][j];
              b[i][j] = 0;
         }
    }
 }
 for(int i = 0; i < n; i++) {
     for(int j = 0; j < n; j++) {
         cout << t[i][j] << " ";
     }
     cout << endl;
 }
 return 0; 
}