线性代数（矩阵乘法）：POJ 3233 Matrix Power Series

Matrix Power Series

 

Description

Given a n × n matrix A and a positive integer k, find the sum S = A + A² + A³ + … + A^k.

Input

The input contains exactly one test case. The first line of input contains three positive integers n (n ≤ 30), k (k ≤ 10⁹) and m (m < 10⁴). Then follow n lines each containing n nonnegative integers below 32,768, giving A’s elements in row-major order.

Output

Output the elements of S modulo m in the same way as A is given.

Sample Input

2 2 4
0 1
1 1

Sample Output

1 2
2 3

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#include <iostream>
#include <cstring>
#include <cstdio>
using namespace std;
const int maxn=50;
int mod,n,K;
struct Matrix{
    int mat[maxn][maxn];
    Matrix(){
        memset(mat,0,sizeof(mat));
    }
    
    Matrix operator +(Matrix a){
        Matrix r;
        for(int i=1;i<=n;i++)
            for(int j=1;j<=n;j++)
                (r.mat[i][j]=(mat[i][j]+a.mat[i][j])%mod)%=mod;
        return r;        
    }
    
    Matrix operator *(Matrix a){
        Matrix r;
        for(int i=1,s;i<=n;i++)
            for(int k=1;k<=n;k++){
                s=mat[i][k];
                for(int j=1;j<=n;j++)
                    (r.mat[i][j]+=(s*a.mat[k][j])%mod)%=mod;        
            }
        return r;    
    }
    
    Matrix operator ^(int k){
        Matrix r,x;
        for(int i=1;i<=n;i++)
            r.mat[i][i]=1;
        for(int i=1;i<=n;i++)
            for(int j=1;j<=n;j++)
                x.mat[i][j]=mat[i][j];
        while(k){
            if(k&1)
                r=r*x;
            k>>=1;
            x=x*x;    
        }        
        return r;
    }
}A,B,ans;
Matrix Solve(int k){
    if(k==1)return A;
    if(k%2)return A+A*Solve(k-1);
    else return ((A^(k/2))+B)*Solve(k/2);
}
int main(){
#ifndef ONLINE_JUDGE
    //freopen("","r",stdin);
    //freopen("","w",stdout);    
#endif
    scanf("%d%d%d",&n,&K,&mod);
    for(int i=1;i<=n;i++)
        for(int j=1;j<=n;j++)
            scanf("%d",&A.mat[i][j]);
    
    for(int i=1;i<=n;i++)
        B.mat[i][i]=1;
    
    ans=Solve(K);
    
    for(int i=1;i<=n;i++){
        for(int j=1;j<=n;j++)
            printf("%d ",ans.mat[i][j]);
        printf("
");    
    }
    return 0;
}