HDU 4920 Matrix multiplication（矩阵乘法小技巧）——2014 Multi-University Training Contest 5

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Matrix multiplication

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 4338    Accepted Submission(s): 1736

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

1 
0 
1 
2 
0 1 
2 3 
4 5 
6 7

 

Sample Output

0 
0 1 
2 1

题目大意：

就是求两个 n∗n 的矩阵相乘。

解题思路：

这个题 正常写的话 应该不会过 肯定会超时的，所以要将矩阵进行转置这样就会过了，我还有一个问题（问题在代码中有提到），如果有人知道 欢迎来帮我解决！！！ /叩谢

My Code：

/**2016 - 09 - 15 下午Author: ITAKMotto:今日的我要超越昨日的我，明日的我要胜过今日的我，以创作出更好的代码为目标，不断地超越自己。**/#include <iostream>#include <cstdio>#include <cstring>#include <cstdlib>#include <cmath>#include <vector>#include <queue>#include <algorithm>#include <set>using namespace std;typedef long long LL;typedef unsigned long long ULL;const int INF = 1e9+5;const int MAXN = 800+5;const int MOD = 3;const double eps = 1e-7;const double PI = acos(-1);using namespace std;int Scan_Int()///输入外挂{    int res=0,ch,flag=0;    if((ch=getchar())=='-')        flag=1;    else if(ch>='0'&&ch<='9')        res=ch-'0';    while((ch=getchar())>='0'&&ch<='9')        res=res*10+ch-'0';    return flag?-res:res;}LL Scan_LL()///输入外挂{    LL res=0,ch,flag=0;    if((ch=getchar())=='-')        flag=1;    else if(ch>='0'&&ch<='9')        res=ch-'0';    while((ch=getchar())>='0'&&ch<='9')        res=res*10+ch-'0';    return flag?-res:res;}void Out(int a)///输出外挂{    if(a>9)        Out(a/10);    putchar(a%10+'0');}int a[MAXN][MAXN], b[MAXN][MAXN];int main(){    int n;    while(~scanf("%d",&n))    {        for(int i=0; i<n; i++)            for(int j=0; j<n; j++)                scanf("%d",&a[i][j]), a[i][j] = a[i][j] % MOD;        /// 矩阵乘法 一定要转置        for(int i=0; i<n; i++)            for(int j=0; j<n; j++)                scanf("%d",&b[j][i]), b[j][i] = b[j][i] % MOD;        for(int i=0; i<n; i++)        {            for(int j=0; j<n; j++)            {                int tmp = 0;                for(int k=0; k<n; k++)                    ///if(a[i][k] && b[j][k]) 为什么加上这句话不过，我很蒙逼                        tmp += a[i][k]*b[j][k];                if(j == 0)                    printf("%d",tmp%MOD);                else                    printf(" %d",tmp%MOD);            }            puts("");        }    }    return 0;}