HDU 4920 Matrix multiplication

Matrix multiplication

Time Limit: 2000ms

Memory Limit: 131072KB

This problem will be judged on HDU. Original ID: 4920
64-bit integer IO format: %I64d      Java class name: Main

 

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

Source

2014 Multi-University Training Contest 5

 

解题：利用cache进行加速。。

 

#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
const int maxn = 810;
int a[maxn][maxn],b[maxn][maxn],c[maxn][maxn];
int main() {
    int n;
    while(~scanf("%d",&n)) {
        memset(c,0,sizeof c);
        for(int i = 0; i < n; ++i)
            for(int j = 0; j < n; ++j) {
                scanf("%d",a[i]+j);
                a[i][j] %= 3;
            }
        for(int i = 0; i < n; ++i)
            for(int j = 0; j < n; ++j) {
                scanf("%d",b[i]+j);
                b[i][j] %= 3;
            }
        for(int i = 0; i < n; ++i)
            for(int k = 0; k < n; ++k)
                for(int j = 0; j < n; ++j)
                    c[i][j] += a[i][k]*b[k][j];
        for(int i = 0; i < n; ++i) {
            for(int j = 0; j + 1 < n; ++j)
                printf("%d ",c[i][j]%3);
            printf("%d
",c[i][n-1]%3);
        }
    }
    return 0;
}