poj3233Matrix Power Series(矩阵乘法)

Matrix Power Series

Time Limit: 3000MS		Memory Limit: 131072K
Total Submissions: 23187		Accepted: 9662

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

/*
矩阵乘法经典+二分
Sk=A+A2+A3+...+Ak
  =(1+Ak/2)*(A+A2+A3+...+Ak/2)+{Ak}
  =(1+Ak/2)*(Sk/2)+{Ak}
当k为偶数时不要大括号里面的数
*/
#include<iostream>
#include<cstdio>
#include<cstring>

using namespace std;
int n,m,k;
struct matrix
{
    int a[][];
    void init()
    {
        memset(a,,sizeof a);
        for(int i=;i<;i++) a[i][i]=;
    }
};

void print(matrix s)
{
    for(int i=;i<n;i++)
    {
        for(int j=;j<n;j++)
        {
            if (j)
                printf(" ");
            printf("%d",s.a[i][j]%m);
        }
        printf("\n");
    }
}

matrix m_add(matrix a,matrix b)//加法
{
    matrix c;
    for(int i=;i<n;i++)
      for(int j=;j<n;j++)
        c.a[i][j]=((a.a[i][j]+b.a[i][j])%m);
    return c;
}

matrix m_mul(matrix a,matrix b)//乘法
{
    matrix c;
    for(int i=;i<n;i++)
    {
        for(int j=;j<n;j++)
        {
            c.a[i][j]=;
            for(int k=;k<n;k++)
              c.a[i][j]+=((a.a[i][k]*b.a[k][j])%m);
            c.a[i][j]%=m;
        }
    }
    return c;
}

matrix mul(matrix s,int k)//矩阵快速幂
{
    matrix ans;ans.init();
    while(k>=)
    {
        if(k&) ans=m_mul(ans,s);
        k>>=;
        s=m_mul(s,s);
    }
    return ans;
}

matrix sum(matrix s,int k)//矩阵前k项求和
{
    if(k==) return s;
    matrix tmp;tmp.init();
    tmp=m_add(tmp,mul(s,k>>));//计算1+A^(k/2)
    tmp=m_mul(tmp,sum(s,k>>));//计算(1+A^(k/2))*(A+A^2+A^3+...+A^(k/2))
    if(k&) tmp=m_add(tmp,mul(s,k));
    return tmp;
}

int main()
{
    while(cin>>n>>k>>m)
    {
        matrix s;
        for(int i=;i<n;i++)
          for(int j=;j<n;j++)
            scanf("%d",&s.a[i][j]);
        s=sum(s,k);
        print(s);
    }
}