HDU 4920 矩阵乘积 优化

Matrix multiplication

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 1289    Accepted Submission(s): 568

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

 

#include <iostream>
#include <string.h>
#include <stdio.h>

using namespace std;
#define MAX 810
int a[MAX][MAX],b[MAX][MAX],c[MAX][MAX];

int main()
{
    int n;
    while(~scanf("%d",&n))
    {
        for(int i=;i<n;i++)
        {
            for(int j=;j<n;j++)
            {
                scanf("%d",&a[i][j]);
                a[i][j]=a[i][j]%;
            }
        }
        for(int i=;i<n;i++)
        {
            for(int j=;j<n;j++)
            {
                scanf("%d",&b[i][j]);
                b[i][j]=b[i][j]%;
            }
        }

        memset(c,,sizeof(c));
        for(int i=;i<n;i++)
        {
            for(int j=;j<n;j++)
            {
                if(a[i][j]==)  continue;
                for(int k=;k<n;k++)
                {
                    c[i][k]+=a[i][j]*b[j][k];   如果这里 c[i][k]+=(a[i][j]*b[j][k])%3;就会超时
                }
            }
        }
        for(int i = ; i < n; i++)
         {
             for(int j = ; j < n; j++)
             if(j == )
             printf("%d", c[i][j]%);
             else
             printf(" %d", c[i][j]%);
             printf("\n");
         }
    }
    return ;
}

#include <iostream>
#include<stdio.h>
#include<vector>
///他把所有的0都忽略了，很巧妙的优化，aa[][], bb[][]里存储的是下一个不为0的位置：
#include<queue>
#include<stack>
#include<string.h>
#include<algorithm>
#include<map>
using namespace std;
#define LL long long
#define gcd(a,b) (b==0?a:gcd(b,a%b))
#define lcm(a,b) (a*b/gcd(a,b))
//O（n）求素数，1-n的欧拉数
#define N 100010
//A^x = A^(x % Phi(C) + Phi(C)) (mod C)
int a[][];
int aa[][];
int b[][];
int bb[][];

int c[][];
int main()
{
    int n;
    while(~scanf("%d",&n))
    {
        memset(a,,sizeof(a));
        memset(b,,sizeof(b));
        memset(aa,,sizeof(aa));
        memset(bb,,sizeof(bb));
        memset(c,,sizeof(c));
        for(int i=; i<=n; i++)
        {
            for(int j=; j<=n; j++)
            {
                scanf("%d",&a[i][j]);
                a[i][j]%=;
               /// cout<<"%%%"<<a[i][j]<<endl;
            }
        }
        for(int i=; i<=n; i++)
        {
            for(int j=; j<=n; j++)
            {
                scanf("%d",&b[i][j]);
                b[i][j]%=;
            }
        }
        for(int i=; i<=n; i++)
        {
            int x=-;
            for(int j=n; j>=; j--)
            {
                aa[i][j]=x;    ///后退了一位  所以j：n~~0
                if(a[i][j]) x=j;
                //cout<<"~~~~"<<a[i][j]<<' '<<aa[i][j]<<endl;///记录矩阵中0的位置  赋值给aa【】【】
            }
        }
        for(int i=; i<=n; i++)
        {
            int x=-;
            for(int j=n; j>=; j--)
            {
                bb[i][j]=x;
                if(b[i][j])x=j;
            }
        }
        for(int i=; i<=n; i++)
        {
            for(int j=aa[i][]; j!=-; j=aa[i][j])
            {
                for(int k=bb[j][]; k!=-; k=bb[j][k])  ///a==0时 对应b那一列乘a==0   所以根据bb【j】来转变
                    c[i][k]+=a[i][j]*b[j][k];
            }
        }
        for(int i=; i<=n; i++)
        {
            for(int j=; j<=n; j++)
            {
                printf("%d",c[i][j]%);
                if(j!=n)printf(" ");
                else printf("\n");
            }
        }
    }
    return ;
}