POJ 3233 Matrix Power Series （矩阵分块，递推）

矩阵乘法是可以分块的，而且幂的和也是具有线性的。

不难得到 S_i = S_i-1+A*A_i-1，A_i = A*A_i-1。然后矩阵快速幂就可以了。

/*********************************************************
*            ------------------                          *
*   author AbyssalFish                                   *
**********************************************************/
#include<cstdio>
#include<iostream>
#include<string>
#include<cstring>
#include<queue>
#include<vector>
#include<stack>
#include<vector>
#include<map>
#include<set>
#include<algorithm>
#include<cmath>
#include<ctime>
using namespace std;

typedef long long ll;
typedef vector<int> row;
typedef vector<row> mat;

int n, k, M;

mat Mul;
mat &operator *(mat &A, mat& B)
{
    mat &R = Mul;
    R.assign(n,row(n));
    for(int i = ; i < n; i++){
        for(int j = ; j < n; j++){
            for(int k = ; k < n; k++){
                R[i][j] = (R[i][j] +A[i][k]*B[k][j])%M;

            }
        }
    }

    return R;
}

//#define LOCAL
#ifdef LOCAL
void censor(mat &B)
{
    for(auto r: B){
        for(int c: r)
            cout<<c<<' ';
        cout<<endl;
    }
}
#endif

mat operator ^(mat A,int q)
{
    mat Re(n,row(n));
    for(int i = ; i < n; i++) Re[i][i] = ;
    while(q){
        if(q&) Re = Re*A;
        A = A*A;
        q >>= ;
    }
    return Re;
}

int main()
{
#ifdef LOCAL
    freopen("in.txt","r",stdin);
#endif
    int nn; scanf("%d%d%d",&nn,&k,&M);
    n = *nn;
    mat A(nn,row(nn));
    for(int i = ; i < nn; i++){
        for(int j = ; j < nn; j++){
            scanf("%d",&A[i][j]);
        }
    }
    mat B(n,row(n));
    for(int i = ; i < nn; i++) {
        B[i][i] = ;
        copy(A[i].begin(),A[i].end(),B[i].begin()+nn);
        copy(A[i].begin(),A[i].end(),B[i+nn].begin()+nn);
    }
    B = B^k;
    for(int i = ; i < nn; i++){
        for(int j = ; j < nn; j++){
            printf("%d%c",B[i][j+nn],j==nn-?'\n':' ');
        }
    }
#ifdef LOCAL
    cout<<"rum time:"<<clock()<<"ms"<<endl;
#endif // LOCAL
    return ;
}