hdu 4920 矩阵循环

Matrix multiplication

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)

Total Submission(s): 3273    Accepted Submission(s): 1382


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 Aij. The next n lines describe the matrix B in similar format (0≤Aij,Bij≤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 <cstdio>
#include <cstring>

using namespace std;

#define maxn 805
typedef long long ll;
int a[maxn][maxn];
int b[maxn][maxn];
int c[maxn][maxn];
int n;

int main()
{
while(~scanf("%d", &n)){
for(int i = 1; i <= n; i++)
for(int j = 1; j <= n; j++){
scanf("%lld", &a[i][j]);
a[i][j] %= 3;
}
for(int i = 1; i <= n; i++)
for(int j = 1; j <= n; j++){
scanf("%lld", &b[i][j]);
b[i][j] %= 3;
}

memset(c, 0, sizeof(c));
for(int i = 1; i <= n; i++)
for(int j = 1; j <= n; j++)
for(int k = 1; k <= n; k++) //注意内层循环顺序，利用cpu cache机制
c[i][k]=(c[i][k]+a[i][j]*b[j][k]);

for(int i = 1; i <= n; i++){
for(int j = 1; j <= n; j++){
printf("%d", c[i][j]%3);
if(j < n)
printf(" ");
else
printf("\n");
}
}

}
return 0;
}