Power of Matrix（uva11149+矩阵快速幂）

Power of Matrix

Time Limit:3000MS     Memory Limit:0KB     64bit IO Format:%lld & %llu

Submit Status Practice UVA 11149

Appoint description: 
System Crawler  (2015-03-15)

Description

 

Problem B : Power of Matrix

Time limit: 10 seconds

Consider an n-by-n matrix A. We define A^k = A * A * ... * A (k times). Here, * denotes the usual matrix multiplication.

You are 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 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.

Output

For each case, your program should compute the matrix A + A² + A³ + ... + A^k. Since the values may be very large, you only need to print their last digit. Print a blank line after each case.

Sample Input

3 2
0 2 0
0 0 2
0 0 0
0 0

Sample Output

0 2 4
0 0 2
0 0 0

首先我们来想一下计算A+A^2+A^3...+A^k。

如果A=2,k=6。那你怎么算

2+2²+2³+2⁴+2⁵+2⁶ = ？= (2+2²+2³)*(1+2³)

如果A=2,k=7。那你怎么算

2+2²+2³+2⁴+2⁵+2⁶+2⁷ = ？= (2+2²+2³)*(1+2³)+2⁷

^{so....同理：}

当k是偶数，A+A^2+A^3...+A^k=（E+A^(k/2)）*(A+A^2...+A^(k/2))。

当k是奇数，A+A^2+A^3...+A^k=（E+A^(k/2)）*(A+A^2...+A^(k/2))+A^k。

转载请注明出处：寻找&星空の孩子

题目链接：UVA 11149

#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,,sizeof(c.mat));
    for(int i=; i<N; i++)
    {
        for(int j=; j<N; j++)
        {
            if(a.mat[i][j]==)continue;
            for(int k=; k<N; k++)
            {
                if(b.mat[j][k]==)continue;
                c.mat[i][k]=(c.mat[i][k]+a.mat[i][j]*b.mat[j][k])%;
 
            }
        }
    }
    return c;
}
 
matrix quickmod(matrix a,int n)
{
    matrix res;
    for(int i=; i<N; i++) //单位阵
        for(int j=; j<N; j++)
            res.mat[i][j]=(i==j);
    while(n)
    {
        if(n&)
            res=multiply(a,res);
        a=multiply(a,a);
        n>>=;
    }
    return res;
}
matrix add (matrix a,matrix b)
{
    matrix ret;
    for(int i=; i<N; i++)
        for(int j=; j<N; j++)
        ret.mat[i][j]=(a.mat[i][j]+b.mat[i][j])%;
    return ret;
}
matrix solve(matrix a,int k)
{
    if(k==) return a;
    matrix ans;
    for(int i=; i<N; i++)
        for(int j=; j<N; j++)
            ans.mat[i][j]=(i==j);
    if(k==) return ans;
    ans=multiply((add(quickmod(a,(k>>)),ans)),solve(a,(k>>)));
    if(k%) 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=;i<N;i++)
        {
            for(int j=;j<N;j++)
            {
                int temp;
                scanf("%d",&temp);
                ans.mat[i][j]=temp%;
            }
        }
 
        ans=solve(ans,k);
 
        for(int i=;i<N;i++)
        {
            for(int j=;j<N-;j++)
            {
                printf("%d ",ans.mat[i][j]);
            }
            printf("%d\n",ans.mat[i][N-]);
        }
        printf("\n");
    }
    return ;
}