poj 3233 Matrix Power Series（二分）

Matrix Power Series

Time Limit: 3000MS Memory Limit: 131072KTotal Submissions: 14827 Accepted: 6365

Description

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

Input

The input contains exactly one test case. The first line of input contains three positive integersn (n ≤ 30), k (k ≤ 10⁹) and m (m < 10⁴). Then follown lines each containing n nonnegative integers below 32,768, givingA’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 40 11 1

Sample Output

Sample Output

1 22 3

题目描述：求矩阵幂的和。思路：用到了二分的思想，当k为偶数时S=A+A^2+...+A^(k/2)+A^(k/2)(A+A^2+...+A^(k/2))，当k为奇数时S=A+A^2+...+A^(k/2)+A^(k/2)(A+A^2+...+A^(k/2))+A^k，不断分下去，代码WA了好几次，原因是对取模运算不熟悉，这里做一个整理。用C++写了一个矩阵类，代码如下：

/***************************** (a+b)%mod=(a%mod+b%mod)%mod (a-b)%mod=(a%mod-b%mod)%mod (a*b)%mod=(a%mod*b%mod)%mod a^b%mod=((a%mod)^b)%mod******************************/#include<iostream>#include<stdio.h>using namespace std;class Matrix{public:    Matrix(int s);    Matrix operator+(const Matrix& b);    Matrix operator%(int mod);    Matrix operator*(const Matrix& b);    friend istream& operator>>(istream& in,Matrix& a);    friend ostream& operator<<(ostream& out,Matrix& a);    void Init();    int n,**A;};void Matrix::Init(){    for(int i=0;i<n;i++)        for(int j=0;j<n;j++)A[i][j]=1;}Matrix::Matrix(int s){    n=s;    int i,j;    A=new int*[s];    for(i=0; i<s; i++)A[i]=new int[s];}Matrix Matrix::operator+(const Matrix& b){    Matrix tmp(b.n);    int i,j;    for(i=0; i<b.n; i++)        for(j=0; j<b.n; j++)            tmp.A[i][j]=A[i][j]+b.A[i][j];    return tmp;}Matrix Matrix::operator%(int mod){    for(int i=0; i<n; i++)        for(int j=0; j<n; j++)            A[i][j]%=mod;    return *this;}Matrix Matrix::operator*(const Matrix& b){    int i,j;    Matrix tmp(n);    for(i=0; i<n; i++)        for(j=0; j<n; j++)        {            int sum=0;            for(int di=0; di<n; di++)                sum+=A[i][di]*b.A[di][j];            tmp.A[i][j]=sum;        }    return tmp;}istream& operator>>(istream& in,Matrix &a){    int i,j;    for(i=0;i<a.n;i++)        for(j=0;j<a.n;j++)            in>>a.A[i][j];    return in;}ostream& operator<<(ostream& out,Matrix& a){    int i,j;    for(i=0;i<a.n;i++)    {        for(j=0;j<a.n;j++)            out<<a.A[i][j]<<' ';        out<<endl;    }    return out;}Matrix pow(Matrix a,int b,int mod){    if(b==1)return a%mod;    Matrix tmp=pow(a,b/2,mod);    if(b&0x1)return ((tmp%mod*(tmp%mod))%mod*(a%mod))%mod;    return (tmp%mod*(tmp%mod))%mod;}Matrix Sum(Matrix a,int b,int mod){    if(b==1)return a%mod;    Matrix tmp=Sum(a,b/2,mod);    Matrix t=pow(a,b/2,mod);    if(!(b&0x1))return (tmp%mod+(t%mod)*(tmp%mod))%mod;    return (tmp%mod+(t%mod*(tmp%mod))%mod+pow(a,b,mod)%mod)%mod;}int main(){    int n,k,m;    cin>>n>>k>>m;    Matrix a(n);    cin>>a;    Matrix ans=Sum(a,k,m);    cout<<ans;    return 0;}