hdu 4920

http://acm.hdu.edu.cn/showproblem.php?pid=4920

给定两个n阶矩阵，求矩阵相乘后模3.

直接搞肯定会超时

特殊处理1和2的情况

实际上是水过的.....

貌似bitset这样的可搞

http://blog.csdn.net/keshuai19940722/article/details/38391913

#include <cstdlib>
#include <iostream>

using namespace std;

 int a[801][801],b[801][801],c[801][801],b1[801][801];
int main()
{
    int i,j,k,n,temp,ans;
   
    
   while ((scanf("%d",&n))!= EOF )
    {
    
          for (i=1;i<=n;i++)
             for (j=1;j<=n;j++)
           {
               scanf("%d",&temp);
               a[i][j]=temp%3;
           }
          for (i=1;i<=n;i++)
             for (j=1;j<=n;j++)
           {
               scanf("%d",&temp);
               b[i][j]=temp%3;
               b1[i][j]=(temp*2)%3;
           }
                
          /* for (i=1;i<=n;i++)
               for(j=1;j<=n;j++)
                  { 
                     ans = 0;
                     for(k=1;k<=n;k++)
                         ans+=a[i][k]*b[k][j];
                     ans = ans % 3;
                     printf("%d ",ans); 
                     if (j==n) printf("
");
                  }*/
          memset(c,0,sizeof(c));
          for (i=1;i<=n;i++)
              for (j=1;j<=n;j++)
              {
                  if (a[i][j]==1)  
                        for (k=1;k<=n;k++) c[i][k]+=b[j][k];
                  else if (a[i][j]==2)  
                        for (k=1;k<=n;k++) c[i][k]+=b1[j][k];
              }
          for (i=1;i<=n;i++)
          {
              for (j=1;j<=n;j++)
                if (j!=n) printf("%d ",c[i][j]%3);
                else printf("%d",c[i][j]%3);
              printf("
");
          }
          
          
    }
    
    return 0;
   
}