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HDU-4920 Matrix multiplication

2024-07-16 02:52:56   220人阅读

            矩阵相乘，采用一行的去访问，比采用一列访问时间更短，根据数组是一行去储存的。神奇小代码。         

             Matrix multiplication

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 1476    Accepted Submission(s): 650


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
#include<iostream>#include<cstring>#include<cstdio>using namespace std;int a[805][805],b[805][805],c[805][805];int main(){    int k,i,n,j,x;    while(~scanf("%d",&n))    {        memset(c,0,sizeof(c));        for(i=1; i<=n; i++)            for(j=1; j<=n; j++)            {                scanf("%d",&x);                a[i][j]=x%3;            }        for(i=1; i<=n; i++)            for(j=1; j<=n; j++)            {                scanf("%d",&x);                b[i][j]=x%3;            }        for(k=1; k<=n; k++)        {            for(j=1;j<=n; j++)            {                for(i=1;i<=n; i++)                {                    c[k][i]+=a[k][j]*b[j][i];                }            }        }        for(i=1;i<=n; i++)        {            for(j=1;j<n; j++)                printf("%d ",c[i][j]%3);            printf("%d\n",c[i][n]%3);        }    }    return 0;}

 


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