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=;
 int mod,n,K;
 struct Matrix{
     int mat[maxn][maxn];
     Matrix(){
         memset(mat,,sizeof(mat));
     }

     Matrix operator +(Matrix a){
         Matrix r;
         for(int i=;i<=n;i++)
             for(int j=;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=,s;i<=n;i++)
             for(int k=;k<=n;k++){
                 s=mat[i][k];
                 for(int j=;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=;i<=n;i++)
             r.mat[i][i]=;
         for(int i=;i<=n;i++)
             for(int j=;j<=n;j++)
                 x.mat[i][j]=mat[i][j];
         while(k){
             if(k&)
                 r=r*x;
             k>>=;
             x=x*x;
         }
         return r;
     }
 }A,B,ans;
 Matrix Solve(int k){
     if(k==)return A;
     if(k%)return A+A*Solve(k-);
     else return ((A^(k/))+B)*Solve(k/);
 }
 int main(){
 #ifndef ONLINE_JUDGE
     //freopen("","r",stdin);
     //freopen("","w",stdout);
 #endif
     scanf("%d%d%d",&n,&K,&mod);
     for(int i=;i<=n;i++)
         for(int j=;j<=n;j++)
             scanf("%d",&A.mat[i][j]);

     for(int i=;i<=n;i++)
         B.mat[i][i]=;

     ans=Solve(K);

     for(int i=;i<=n;i++){
         for(int j=;j<=n;j++)
             printf("%d ",ans.mat[i][j]);
         printf("\n");
     }
     return ;
 }