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≤10 ⁹).

 

 

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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 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 }