UVA 11149 Power of Matrix

COnsider an n-by-n Matrix A. We define ak = a * a * ... * a (K times). Here, * denotes the usual matrix multiplication.

You is to write a program that computes the matrix a + a2 + a3 + ... + a K.

Example

Suppose A =. Then A2 = =, thus:

Such computation has various applications. For instance, the above example actually counts all the paths in the following graph:

Input

Input consists of no more than test cases. The first line is contains, positive integers n (≤40) and K (≤1000000). This was followed by n lines, each containing n non-negative integers, giving the matrix A.

Input is terminated by a case where n = 0. This case is need not being processed.

Output

For each case, your program should compute the matrix a + a2 + a3 + ... + a k. Since The values may is very large, you have need to print their last digit. Print a blank line after each case.

Sample Input

3 20 2 00 0 20 0 00 0

Sample Output

0 2 40 0 20 0 0


Note that K is even when there is A1 + a^2 + a^3 + .... A^k = (E + a^ (K/2) (a^1+a^2+a^3+... A ^ k/2)
To K is odd when the last think A^k alone to propose the remaining as an even
DFS can

#include <iostream> #include <cstring> #include <cstdio> #include <cstdlib>using namespace std; #define MAXN 50int n,k;struct node{int mat[maxn][maxn];};    Node CALCU (node x, node Y) {node ret;    memset (ret.mat,0,sizeof (Ret.mat));            for (int i = 1, i <= N; i++) for (int j = 1; J <= N; j + +) {for (int k = 1; k <= N; k++)    RET.MAT[I][J] = (Ret.mat[i][j] + x.mat[i][k] * y.mat[k][j])% 10; } return ret;}    Node Add (node X,node y) {node ret;    memset (ret.mat,0,sizeof (Ret.mat)); for (int i = 1, i <= N; i++) for (int j = 1; J <= N; j + +) {Ret.mat[i][j] = X.mat[i][j] + Y.mat I           [j];   RET.MAT[I][J]%= 10; } return ret;}    Node Pow_mat (node X,int cnt) {node ret;    memset (ret.mat,0,sizeof (Ret.mat));    for (int i = 1; i < MAXN; i++) ret.mat[i][i] = 1;        while (CNT) {if (CNT & 1) ret = CALCU (ret,x);        x = CALCU (x,x);    CNT >>= 1; } return ret;} Node DFS (node cur, int k) {if (k = = 1) return cur;    Node res = DFS (CUR,K/2);    Node ans;    Ans = Add (RES,CALCU (Res,pow_mat (CUR,K/2)));    if (K & 1) ans = Add (Ans,pow_mat (cur,k)); return ans;}        int main () {while (scanf ("%d%d", &n,&k)! = EOF) {if (n = = 0) break;        Node ans; for (int i = 1; I <= n; i++) for (int j = 1; J <= N; j + +) {scanf ("%d", &ans.mat[i][j]); ans.mat[        I][J]%= 10;}        node ret = DFS (ans,k);                for (int i = 1; I <= n; i++) {printf ("%d", ret.mat[i][1]);                for (int j = 2; J <= N; j + +) printf ("%d", ret.mat[i][j]);        Putchar (' \ n ');    } putchar (' \ n '); } return 0;}

　　

UVA 11149 Power of Matrix