hdu 4920 Matrix multiplication（矩阵相乘）2014多校训练第5场

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Matrix multiplication

Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)

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

题意：给出两个n*n的矩阵，求这两个矩阵的乘积，结果对3取余。

分析：拿到题先用了经典的矩阵相乘的方法，提交以后果断超时了。后来在网上搜了一下矩阵相乘优化，找到了一个优化方法，只可惜现在我还没有理解是怎么优化的。

#include<cstdio>#include<cstring>#include<algorithm>using namespace std;const int N = 805;int a[N][N], b[N][N], ans[N][N];void  Multi(int n){    int  i, j, k, L, *p2;    int  tmp[N], con;    for(i = 0; i < n; ++i)    {        memset(tmp, 0, sizeof(tmp));        for(k = 0, L = (n & ~15); k < L; ++k)        {            con = a[i][k];            for(j = 0, p2 = b[k]; j < n; ++j, ++p2)                tmp[j] += con * (*p2);            if((k & 15) == 15)            {                for(j = 0; j < n; ++j) tmp[j] %= 3;            }        }        for( ; k < n; ++k)        {            con = a[i][k];            for(j = 0, p2 = b[k]; j < n; ++j, ++p2)                tmp[j] += con * (*p2);        }        for(j = 0; j < n; ++j)            ans[i][j] = tmp[j] % 3;    }}int main(){    int n, i, j, k;    while(~scanf("%d",&n))    {        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;            }        Multi(n);        for(i = 0; i < n; i++)        {            for(j = 0; j < n-1; j++)                printf("%d ", ans[i][j]);            printf("%d\n", ans[i][n-1]);        }    }    return 0;}

http://blog.csdn.net/gogdizzy/article/details/9003369这里面讲解了矩阵相乘的优化方法。

下面这种方法也可以过：

#include<cstdio>#include<cstring>#include<algorithm>#include<cmath>using namespace std;const int N = 805;int a[N][N], b[N][N], ans[N][N];int main(){    int n, i, j, k;    while(~scanf("%d",&n))    {        for(i = 1; i <= n; i++)            for(j = 1; j <= n; j++)            {                scanf("%d",&a[i][j]);                a[i][j] %= 3;            }        for(i = 1; i <= n; i++)            for(j = 1; j <= n; j++)            {                scanf("%d",&b[i][j]);                b[i][j] %= 3;            }        memset(ans, 0, sizeof(ans));        for(k = 1; k <= n; k++) //经典算法中这层循环在最内层，放最内层会超时，但是放在最外层或者中间都不会超时，不知道为什么            for(i = 1; i <= n; i++)                for(j = 1; j <= n; j++)                {                    ans[i][j] += a[i][k] * b[k][j];                    //ans[i][j] %= 3;   //如果在这里对3取余，就超时了                }        for(i = 1; i <= n; i++)        {            for(j = 1; j < n; j++)                printf("%d ", ans[i][j] % 3);            printf("%d\n", ans[i][n] % 3);        }    }    return 0;}

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