矩陣乘法 --- hdu 4920 ： Matrix multiplication

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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 DescriptionGiven 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. 

 

InputThe 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). 

 

OutputFor each tests:

Print n lines. Each of them contain n integers -- the matrix A×B in similar format. 

 

Sample Input10120 12 34 56 7 

 

Sample Output00 12 1 

 

AuthorXiaoxu Guo (ftiasch) 

 

Source2014 Multi-University Training Contest 5  

 

Mean:

給你兩個矩陣，計算兩個矩陣的積。

 

nalyse:

很多人認為這題用暴力過不了，時間複雜度為O(n^3),800^3=512000000，差不多要接近於10^9了，可能是hdu的評測速度給力吧，再加上這題是單點評測，每個評測點的時限都有2000ms，所以說暴力過了也實屬正常。

 

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;}