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 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
Sample Output
0 0 1 2 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;
}