HDU 4920 Matrix multiplication(矩阵相乘)

各种TEL，233啊。没想到是处理掉0的情况就能够过啊。一直以为会有极端数据。没想到居然是这种啊、、在网上看到了一个AC的奇妙的代码，经典的矩阵乘法，仅仅只是把最内层的枚举，移到外面就过了啊、、、有点不理解啊，复杂度不是一样的吗、、

Matrix multiplication

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 640    Accepted Submission(s): 250

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

 

Author

Xiaoxu Guo (ftiasch)

 

Source

2014 Multi-University Training Contest 5

#include <algorithm>
#include <iostream>
#include <stdlib.h>
#include <string.h>
#include <iomanip>
#include <stdio.h>
#include <string>
#include <queue>
#include <cmath>
#include <stack>
#include <map>
#include <set>
#define eps 1e-12
///#define M 1000100
#define LL __int64
///#define LL long long
///#define INF 0x7ffffff
#define INF 0x3f3f3f3f
#define PI 3.1415926535898
#define zero(x) ((fabs(x)<eps)?
0:x)

using namespace std;


const int maxn = 810;

int a[maxn][maxn];
int b[maxn][maxn];
int c[maxn][maxn];
int aa[maxn][maxn];
int bb[maxn][maxn];

int main()
{
    int n;
    while(cin >>n)
    {
        memset(c, 0, sizeof(c));
        memset(aa, 0, sizeof(aa));
        memset(bb, 0, sizeof(bb));
        for(int i = 1; i <= n; i++)
        {
            for(int j = 1; j <= n; j++)
            {
                scanf("%d",&a[i][j]);
                a[i][j] %= 3;
            }
        }
        for(int i = 1; i <= n; i++)
        {
            for(int j = 1; j <= n; j++)
            {
                scanf("%d",&b[i][j]);
                b[i][j] %= 3;
            }
        }
        for(int i = 1; i <= n; i++)
        {
            int x = -1;
            for(int j = n; j >= 0; j--)
            {
                aa[i][j] = x;
                if(a[i][j]) x = j;
            }
        }

        for(int i = 1; i <= n; i++)
        {
            int x = -1;
            for(int j = n; j >= 0; j--)
            {
                bb[i][j] = x;
                if(b[i][j]) x = j;
            }
        }
        for (int i = 1; i <= n; i++)
        {
            for(int j = aa[i][0]; j != -1; j = aa[i][j])
            {
                for(int k = bb[j][0]; k != -1; k = bb[j][k])
                    c[i][k] += a[i][j]*b[j][k];
            }
        }
        for(int i = 1; i <= n; i++)
        {
            for(int j = 1; j <= n-1; j++)
                printf("%d ",c[i][j]%3);
            printf("%d
",c[i][n]%3);
        }
    }
    return 0;
}

这是看到有人交的AC的代码：

#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，可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;   //假设在这里对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
", ans[i][n] % 3);
        }
    }
    return 0;
}