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
0
1
2
0 1
2 3
4 5
6 7
Sample Output
0
0 1
2 1
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;
}