hdu 4920 Matrix multiplication

Matrix multiplication Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 334    Accepted Submission(s): 112

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

 

题解及代码：

[cpp] view plain copy print ?

1. /* 
2. 签到题，简单的矩阵相乘，注意元素0的处理就不会超时了。 
3. */  
4. #include <iostream>  
5. #include <cstdio>  
6. #include <cstring>  
7. using namespace std;  
8. const int mod=3;  
9. int a[810][810],b[810][810];  
10. int temp[810][810];  
11. void multiple(int n,int p)  
12. {  
13.     int i,j,k;  
14.     for(i=0; i<n; i++)  
15.         for(j=0; j<n; j++)  
16.         {  
17.             if(a[i][j]!=0)  
18.                 for(k=0; k<n; k++)  
19.                     temp[i][k]=(temp[i][k]+a[i][j]*b[j][k])%p;  
20.         }  
21.     for(int i=0;i<n;i++)  
22.     {  
23.         for(int j=0;j<n;j++)  
24.             if(j==0) printf("%d",temp[i][j]);  
25.         else printf(" %d",temp[i][j]);  
26.         puts("");  
27.     }  
28. }  
29.   
30.   
31.   
32. void init(int n)  
33. {  
34.    memset(temp,0,sizeof(temp));  
35.    for(int i=0;i<n;i++)  
36.    {  
37.        for(int j=0;j<n;j++)  
38.        {  
39.           scanf("%d",&a[i][j]);  
40.           a[i][j]%=3;  
41.        }  
42.    }  
43.    for(int i=0;i<n;i++)  
44.    {  
45.        for(int j=0;j<n;j++)  
46.        {  
47.           scanf("%d",&b[i][j]);  
48.           b[i][j]%=3;  
49.        }  
50.    }  
51. }  
52.   
53. int main()  
54. {  
55.     int n;  
56.     while(cin>>n)  
57.     {  
58.         init(n);  
59.         multiple(n,mod);  
60.     }  
61.     return 0;  
62. }