poj 3233 Matrix Power Series

最新推荐文章于 2023-12-30 17:20:46 发布

转载最新推荐文章于 2023-12-30 17:20:46 发布 · 61 阅读

0 ·

CC 4.0 BY-SA版权

原文链接：http://www.cnblogs.com/bo-tao/archive/2012/07/25/2608466.html

这是一道经典的矩阵题：

这个题就是要构造一个矩阵；

A 1 * A 1 ---> A*A A + 1 ..... A^k S(k-1) + 1;

0 1 0 1 0 1 0 1;

View Code

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<algorithm>
#include<cmath>
#include<queue>
#include<set>
#include<cstring>
#include<vector>
#include<string>
#define LL long long
using namespace std;
int n,m;
void multi( int B[][64], int A[][64] )
{
    int G[64][64]= {0};
    for( int i = 0 ; i < n ; i ++)
       for( int j = 0 ; j < n ; j ++ )
            for( int k = 0 ; k < n ; k ++ )
                 G[i][j] += B[i][k] * A[k][j];        
    for( int i = 0 ; i < n ; i ++ )
       for( int j = 0 ; j < n ; j ++ )
            B[i][j] = G[i][j]%m;    
}
int main( )
{
    int k,A[64][64],B[64][64];
    while( scanf( "%d %d %d",&n,&k,&m )==3 )
    {
          memset( A, 0 ,sizeof( A ) );
          memset( B, 0, sizeof( B ) );
          for( int i = 0; i < n ; i ++ )
          {
              for( int j = 0 ; j < n ; j ++ )
              {
                   scanf( "%d",&A[i][j] );
                   A[i][j] %= m;    
              }
              A[i][i+n] = 1;
              A[i+n][i+n] = 1;
              B[i][i] = 1;
              B[i+n][i+n] =1;
             
          }
          n <<= 1;
          for( int i = 0 ; (1<<i)<=k+1 ; i ++ )
          {
               if( (1<<i)&(k+1) ) multi( B  ,A );
               multi( A , A );        
          }
          int t = n;
          n>>=1;
          for( int i = 0 ;i < n  ; i++ )
          {
               B[i][i+n] = (B[i][i+n] - 1 + m)%m;
               for( int j = n ; j < t ; j ++ )
               printf( j==(t-1)?"%d\n":"%d ",B[i][j]);        
          }
    }
    return 0;    
}

转载于:https://www.cnblogs.com/bo-tao/archive/2012/07/25/2608466.html