hdu 4920 矩阵循环

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本文介绍如何计算两个n×n矩阵的乘积，并在模3运算下给出结果。

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 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

Sample Output

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;
}