矩阵乘法 --- hdu 4920 : Matrix multiplication

Matrix multiplication

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 820    Accepted Submission(s): 328

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
 
 

Mean:

给你两个矩阵,计算两个矩阵的积。

nalyse:

很多人认为这题用暴力过不了,时间复杂度为O(n^3),800^3=512000000,差不多要接近于10^9了,可能是hdu的评测速度给力吧,再加上这题是单点评测,每个评测点的时限都有2000ms,所以说暴力过了也实属正常。

Time complexity:O(n^3)

Source code:

#include<stdio.h>
int a[800][800],b[800][800],c[800][800],n,i,j,k;
int main(){
while(scanf("%d",&n)!=EOF){
for(i=0;i<n;++i)
for(j=0;j<n;++j)
scanf("%d",&a[i][j]),a[i][j]%=3;
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<i;++j)
k=b[i][j],b[i][j]=b[j][i],b[j][i]=k;
for(i=0;i<n;++i)
for(j=0;j<n;++j){
c[i][j]=0;
for(k=0;k<n;++k)
c[i][j]+=a[i][k]*b[j][k];
c[i][j]%=3;
}
for(i=0;i<n;++i)
for(j=0;j<n;++j)
printf(j==n-1?"%d\n":"%d ",c[i][j]);
}
return 0;
}

  

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