hdu4920Matrix multiplication(模3矩阵相乘)

Matrix multiplication

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 4275    Accepted Submission(s): 1718

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 A_ij. The next n lines describe the matrix B in similar format (0≤A_ij,B_ij≤10⁹).

 

Output

For each tests:

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

 

Sample Input

10120 12 34 56 7

 

Sample Output

00 12 1

 模3矩阵相乘，摸3后只有0,1,2   3个情况，去掉a[i][k]，为0的情况，相当于减少一维情况，这样floyed就能解了

#include<cstdio>#include<cmath>#include<iostream>#include<cstring>#include<algorithm>#include<cmath>#include<map>#include<set>#include<stack>#include<queue>using namespace std;#define ll __int64#define usint unsigned int#define mz(array) memset(array, 0, sizeof(array))#define minf(array) memset(array, 0x3f, sizeof(array))#define REP(i,n) for(int i=0;i<(n);i++)#define FOR(i,x,n) for(int i=(x);i<=(n);i++)#define RD(x) scanf("%d",&x)#define RD2(x,y) scanf("%d%d",&x,&y)#define RD3(x,y,z) scanf("%d%d%d",&x,&y,&z)#define WN(x) printf("%d\n",x);#define RE  freopen("D.in","r",stdin)#define WE  freopen("1.out","w",stdout)using namespace std;int a[800][800],b[800][800],c[800][800];int n;int main(){    while(scanf("%d",&n)!=EOF)    {        for(int i=0; i<n; i++)            for(int j=0; j<n; j++)            {                scanf("%d",&a[i][j]);                a[i][j]%=3;            }        for(int i=0; i<n; i++)            for(int j=0; j<n; j++)            {                scanf("%d",&b[i][j]);                b[i][j]%=3;            }            memset(c,0,sizeof(c));        for(int k=0; k<n; k++)            for(int i=0; i<n; i++)                for(int j=0; j<n; j++)                    c[i][j]+=a[i][k]*b[k][j];        for(int i=0; i<n; i++)        {            for(int j=0; j<n-1; j++)            {                putchar(c[i][j]%3+'0');                putchar(' ');            }            putchar(c[i][n-1]%3+'0');            puts("");        }    }    return 0;}