HDU 4920 Matrix multiplication (Matrix multiplication), hdumultiplication

HDU 4920 Matrix multiplication (Matrix multiplication), hdumultiplication

Various TEL, 233 ah. I didn't expect it would be enough to deal with 0. I always thought there would be extreme data. I did not expect this. I saw an amazing AC code on the Internet. The classic matrix multiplication only enumerated the innermost layer, it's time to move outside, and it's a bit difficult to understand. Isn't the complexity the same ,,

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 DescriptionGiven 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.
InputThe 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. 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 ).
OutputFor 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

 
AuthorXiaoxu Guo (ftiasch)
Source2014 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\n",c[i][n]%3);        }    }    return 0;}

Here is the code of the AC that someone handed in:

# 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 ++) // in a classic algorithm, this layer of loop exists in the innermost layer, which times out, but not in the outermost layer or in the middle, I do not know why 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; // if the remainder of 3 is obtained here, it will time out} 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 ;}