首页 > 代码库 > HDU4920：Matrix multiplication

HDU4920：Matrix multiplication

2024-07-16 02:04:50   220人阅读

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


对于二维数组，按行取比按列取要快
真心被逗了
不过也算涨了姿势。。。
#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;

int a[805][805],b[805][805],c[805][805];

int main()
{
    int n,i,j,k;
    while(~scanf("%d",&n))
    {
        for(i = 0; i<n; i++)
            for(j = 0; j<n; j++)
            {
                scanf("%d",&a[i][j]);
                a[i][j]%=3;
                c[i][j] = 0;
            }
        for(i = 0; i<n; i++)
            for(j = 0; j<n; j++)
            {
                scanf("%d",&b[i][j]);
                b[i][j]%=3;
            }
        for(i = 0; i<n; i++)
            for(j = 0; j<n; j++)
            {
                if(!a[i][j])
                continue;
                for(k = 0; k<n; k++)
                    c[i][k] = c[i][k]+a[i][j]*b[j][k];
            }
        for(i = 0; i<n; i++)
        {
            for(j = 0; j<n; j++)
                if(j==n-1)
                    printf("%d\n",c[i][j]%3);
                else
                    printf("%d ",c[i][j]%3);
        }
    }

    return 0;
}



联系
我们