HDU 4920 Matrix multiplication 矩阵相乘。稀疏矩阵

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 Description

Given 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.

 

Input

The 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 A_ij. The next n lines describe the matrix B in
similar format (0≤A_ij,B_ij≤10⁹).

 

Output

For each tests:

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

 

Sample Input

1

0

1

2

0 1

2 3

4 5

6 7

 

Sample Output

0

0 1

2 1

 

Author

Xiaoxu Guo (ftiasch)

 

Source

2014
Multi-University Training Contest 5

 

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 #include<iostream>
 #include<stdio.h>
 #include<cstring>
 #include<cstdlib>
 using namespace std;

 struct Matrix
 {
     int mat[][];
 }hxl,tom,now;
 void solve(int n)
 {
     int i,j,k;
     for(i=;i<=n;i++)
     {
         for(k=;k<=n;k++)
         {
             if(hxl.mat[i][k]==) continue;
             for(j=;j<=n;j++)
             {
                 now.mat[i][j] = now.mat[i][j] + hxl.mat[i][k]*tom.mat[k][j];
             }
         }
     }
     for(i=;i<=n;i++)
     {
         for(j=;j<=n;j++)
         {
             if(j==)printf("%d",now.mat[i][j]%);
             else printf(" %d",now.mat[i][j]%);
         }
         printf("\n");
     }
 }
 int main()
 {
     int n;
     int i,j;
     while(scanf("%d",&n)>)
     {
         for(i=;i<=n;i++)
             for(j=;j<=n;j++)
             {
                 scanf("%d",&hxl.mat[i][j]);
                 hxl.mat[i][j]%=;
                 now.mat[i][j] = ;
             }
         for(i=;i<=n;i++)
             for(j=;j<=n;j++)
             {
                 scanf("%d",&tom.mat[i][j]);
                 tom.mat[i][j]%=;
             }
         solve(n);

     }
     return ;
 }