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,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;

}