标签：
|
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 are to write a program that computes the matrix A + A2 + A3 + ... + Ak.
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 consists of no more than 20 test cases. The first line for each case contains two positive integers n (≤ 40) and k (≤ 1000000). This is 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 need NOT be processed.
For each case, your program should compute the matrix A + A2 + A3 + ... + Ak. Since the values may be very large, you only need to print their last digit. Print a blank line after each case.
3 2 0 2 0 0 0 2 0 0 0 0 0
0 2 4 0 0 2 0 0 0
#include<cstdio> #include<cstring> #include<iostream> #include<algorithm> using namespace std; #define LL __int64 #define mmax 45 struct matrix { int mat[mmax][mmax]; }; int N; matrix multiply(matrix a,matrix b) { matrix c; memset(c.mat,0,sizeof(c.mat)); for(int i=0; i<N; i++) { for(int j=0; j<N; j++) { if(a.mat[i][j]==0)continue; for(int k=0; k<N; k++) { if(b.mat[j][k]==0)continue; c.mat[i][k]=(c.mat[i][k]+a.mat[i][j]*b.mat[j][k])%10; } } } return c; } matrix quickmod(matrix a,int n) { matrix res; for(int i=0; i<N; i++) //单位阵 for(int j=0; j<N; j++) res.mat[i][j]=(i==j); while(n) { if(n&1) res=multiply(a,res); a=multiply(a,a); n>>=1; } return res; } matrix add (matrix a,matrix b) { matrix ret; for(int i=0; i<N; i++) for(int j=0; j<N; j++) ret.mat[i][j]=(a.mat[i][j]+b.mat[i][j])%10; return ret; } matrix solve(matrix a,int k) { if(k==1) return a; matrix ans; for(int i=0; i<N; i++) for(int j=0; j<N; j++) ans.mat[i][j]=(i==j); if(k==0) return ans; ans=multiply((add(quickmod(a,(k>>1)),ans)),solve(a,(k>>1))); if(k%2) ans=add(quickmod(a,k),ans); return ans; } int main() { int k; while(scanf("%d%d",&N,&k)!=EOF) { if(!N)break; matrix ans; for(int i=0;i<N;i++) { for(int j=0;j<N;j++) { int temp; scanf("%d",&temp); ans.mat[i][j]=temp%10; } } ans=solve(ans,k); for(int i=0;i<N;i++) { for(int j=0;j<N-1;j++) { printf("%d ",ans.mat[i][j]); } printf("%d\n",ans.mat[i][N-1]); } printf("\n"); } return 0; }
Power of Matrix（uva11149+矩阵快速幂）
标签：
原文地址：http://www.cnblogs.com/yuyixingkong/p/4348677.html
