HDU 4920 Matrix multiplication 矩陣相乘。疏鬆陣列

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Matrix multiplication

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 1775    Accepted Submission(s): 796

Problem DescriptionGiven two matrices A and B of size n×n, find the product of them.

bobo hates big integers. So you are only asked to find the result modulo 3. 

 

InputThe input consists of several tests. For each tests:

The first line contains n (1≤n≤800). Each of the following n lines contain n integers -- the description of the matrix A. The j-th integer in the i-th line equals Aij. The next n lines describe the matrix B in similar format (0≤Aij,Bij≤109). 

 

OutputFor each tests:

Print n lines. Each of them contain n integers -- the matrix A×B in similar format. 

 

Sample Input10120 12 34 56 7 

 

Sample Output00 12 1 AuthorXiaoxu Guo (ftiasch) 

 

Source2014 Multi-University Training Contest 5  

 

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 1 #include<iostream> 2 #include<stdio.h> 3 #include<cstring> 4 #include<cstdlib> 5 using namespace std; 6  7 struct Matrix 8 { 9     int mat[801][801];10 }hxl,tom,now;11 void solve(int n)12 {13     int i,j,k;14     for(i=1;i<=n;i++)15     {16         for(k=1;k<=n;k++)17         {18             if(hxl.mat[i][k]==0) continue;19             for(j=1;j<=n;j++)20             {21                 now.mat[i][j] = now.mat[i][j] + hxl.mat[i][k]*tom.mat[k][j];22             }23         }24     }25     for(i=1;i<=n;i++)26     {27         for(j=1;j<=n;j++)28         {29             if(j==1)printf("%d",now.mat[i][j]%3);30             else printf(" %d",now.mat[i][j]%3);31         }32         printf("\n");33     }34 }35 int main()36 {37     int n;38     int i,j;39     while(scanf("%d",&n)>0)40     {41         for(i=1;i<=n;i++)42             for(j=1;j<=n;j++)43             {44                 scanf("%d",&hxl.mat[i][j]);45                 hxl.mat[i][j]%=3;46                 now.mat[i][j] = 0;47             }48         for(i=1;i<=n;i++)49             for(j=1;j<=n;j++)50             {51                 scanf("%d",&tom.mat[i][j]);52                 tom.mat[i][j]%=3;53             }54         solve(n);55 56     }57     return 0;58 }