HDU 4920 Matrix multiplication 解题报告（暴力）

Matrix multiplication

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 862    Accepted Submission(s): 343

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

 

Author

Xiaoxu Guo (ftiasch)

 

Source

2014 Multi-University Training Contest 5

 
    解题报告：直接乘的复杂度略高，优化一点就可以过了。这里统计按行来计算了，代码如下：

#include <iostream>#include <cstdio>#include <algorithm>#include <cstring>#include <cmath>#include <queue>#include <vector>#include <map>#include <set>#include <string>#include <iomanip>using namespace std;#define ff(i, n) for(int i=0;i<(n);i++)#define fff(i, n, m) for(int i=(n);i<=(m);i++)#define dff(i, n, m) for(int i=(n);i>=(m);i--)#define bit(n) (1LL<<(n))typedef long long LL;typedef unsigned long long ULL;const LL inf=1e15;void work();int main(){#ifdef ACM    freopen("input.in", "r", stdin);//    freopen("input.in", "w", stdout);#endif // ACM    work();}/***************************************************/int a[888][888];int b[888][888];int ans[888];int last[888];int n, tmp;void add(int * x, int * y){    ff(i, n) x[i] += y[i];}void sub(int * x, int * y){    ff(i, n) x[i] -= y[i];}void work(){    while(scanf("%d", &n) == 1)    {        ff(i, n) ff(j, n)            scanf("%d", &tmp), a[i][j] = tmp%3;        ff(i, n) ff(j, n)            scanf("%d", &tmp), b[i][j] = tmp%3;        memset(ans, 0, sizeof(ans));        memset(last, 0, sizeof(last));        ff(i, n)        {            ff(j, n)            {                if(a[i][j] - last[j] == 1 || a[i][j] - last[j] == -2)                    add(ans, b[j]);                else if(a[i][j] - last[j] == -1 || a[i][j] - last[j] == 2)                    sub(ans, b[j]);                last[j] = a[i][j];            }            ff(j, n) ans[j] = (ans[j]%3+3)%3;            printf("%d", ans[0]);            fff(j, 1, n-1) printf(" %d", ans[j]);            puts("");        }    }}