Poj3233 matrix power series matrix multiplication

This question is the third application of matrix multiplication. However, pleaseS=A+A2 +A3 +... +
AK. The result matrix M67 Daniel's blog mentioned above is a sub-governance method. Set f [N] = a ^ 1 + a ^ 2 +... a ^ N;
If n is an even number, F [N] = f [n/2] + F [n/2] * a ^ (n/2 );
However, n is an odd number, and f [N] = f [n-1] + A ^ (N). After reading discuss, we found a method with lower programming complexity. Again ..

Let B = [[a I];
[0 I]

B ^ (k + 1) = [[A ^ k I + A +... + A ^ K];
[0 I]

It's amazing to say that the matrix .......

I am relatively lazy. So... The second method is implemented ....

 

Matrix Power Series

Time limit:3000 Ms		Memory limit:131072 K
Total submissions:11648		Accepted:4977

Description

GivenN×NMatrixAAnd a positive integerK, Find the sumS=
A+A2 +A3 +... +
AK.

Input

The input contains exactly one test case. The first line of input contains three positive integersN(N≤ 30 ),
K(K≤ 109) andM(M<104). Then followNLines each containing
NNonnegative integers below 32,768, givingA'S elements in row-Major Order.

Output

Output the elementsSModuloMIn the same wayAIs given.

Sample Input

2 2 40 11 1

Sample output

1 22 3

#include<iostream>#include<cstdio>#define MAXN 100using namespace std;struct Matrix{    int size;    long modulo;    long element[MAXN][MAXN];    void setSize(int);    void setModulo(int);    Matrix operator* (Matrix);    Matrix power(int);};void Matrix::setSize(int a){    for (int i=0; i<a; i++)        for (int j=0; j<a; j++)            element[i][j]=0;    size = a;}void Matrix::setModulo(int a){    modulo = a;}Matrix Matrix::operator* (Matrix param){    Matrix product;    product.setSize(size);    product.setModulo(modulo);    for (int i=0; i<size; i++)        for (int j=0; j<size; j++)            for (int k=0; k<size; k++)            {                product.element[i][j]+=element[i][k]*param.element[k][j];                product.element[i][j]%=modulo;            }    return product;}Matrix Matrix::power(int exp){    Matrix tmp = (*this) * (*this);    if (exp==1) return *this;    else if (exp & 1) return tmp.power(exp/2) * (*this);    else return tmp.power(exp/2);}int n,m,k;Matrix m1,ans;int main(){    while(~scanf("%d%d%d",&n,&k,&m))    {        m1.setModulo(m);        m1.setSize(n*2);        for(int i=0;i<n;i++)        {            for(int j=0;j<n;j++)            {                scanf("%d",&m1.element[i][j]);                m1.element[i][j]%=m;            }            m1.element[i][i+n]=1;            m1.element[n+i][n+i]=1;        }        m1.element[n][n]=1;        ans=m1.power(k+1);        for(int i=0;i<n;i++)            for(int j=n;j<n*2;j++)            {                int out=ans.element[i][j];                if(j==i+n)                    out=out==0?m-1:out-1;                printf("%d%s",out,j!=n*2-1?" ":"\n");            }        return 0;    }    return 0;}