UVA 11149-power of matrix (equal-to-matrix summation)

Problem B:power of Matrix
Time Limit:10 seconds

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

Your program should compute the matrix a + a2 + a3 + ... + a k< /c12>. 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

Problemsetter:mak Yan Kei

Like the POJ 3233-matrix Power series http://blog.csdn.net/kalilili/article/details/44926947

172 ms #include <cstdio> #include <iostream> #include <cstring> #include <algorithm>using    namespace Std;int n,m;struct mat{int a[45][45];    Mat () {memset (a,0,sizeof (a));    }};mat A;mat i;mat Add (Mat m1,mat m2) {mat ans;    for (int i=1;i<=n;i++) for (int j=1;j<=n;j++) ans.a[i][j]= (m1.a[i][j]+m2.a[i][j])%10; return ans;}    Mat Mul (Mat m1,mat m2) {mat ans;                     for (int i=1;i<=n;i++) for (int. j=1;j<=n;j++) if (M1.a[i][j]) for (int k=1;k<=n;k++)    Ans.a[i][k]= (Ans.a[i][k]+m1.a[i][j]*m2.a[j][k])%10; return ans;}    Mat Quickmul (Mat m,int k) {Mat ans=i;        while (k) {if (k&1) Ans=mul (ans,m);        M=mul (M,M);    k>>=1; } return ans; void print (Mat m) {for (Int. i=1;i<=n;i++) for (int j=1;j<=n;j++) printf ("%d%c", m.a[i][j],j==n? ' \ n ': ');}    Mat getsum (int k) {if (k==1) return A;    Mat Ans=getsum (K/2);   if (k&1) {Mat T=quickmul (a,k/2+1);    Ans=add (Mul (Add (I,t), ans), t);        } else {Mat t=quickmul (A,K/2);    Ans=mul (Add (I,t), ans); } return ans; int main () {while (scanf ("%d%d", &n,&m), N) {for (Int. i=1;i<=n;i++) for (int j=1;j<=n;j+        +) scanf ("%d", &a.a[i][j]), a.a[i][j]%=10;        for (int i=1;i<=n;i++) i.a[i][i]=1;        Mat Ans=getsum (m);        print (ANS);    Puts (""); } return 0;}

UVA 11149-power of matrix (equal-to-matrix summation)