hdu4920Matrix multiplication (矩阵，bitset)

最新推荐文章于 2024-07-31 10:00:00 发布

Kirito_Acmer

最新推荐文章于 2024-07-31 10:00:00 发布

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分类专栏： bitset 文章标签： bitset

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本文介绍了一种矩阵乘法的问题解决方法，包括常规矩阵乘法的优化实现和使用bitset进行高效计算的技巧。

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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

Sample Output

0
0 1
2 1

题意：给你两个矩阵，让你把它们相乘后输出%3后的结果。

思路：普通的矩阵乘法+优化能过的，还有一种思路是开bitset<1000>bt[3],ct[3]，储存每一行mod3后为0,1,2的情况，那么1*1=1,1 *2=2,2*1=2,2*2=1.

代码一：普通矩阵乘法+优化

#include<iostream>
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<math.h>
#include<vector>
#include<map>
#include<set>
#include<queue>
#include<stack>
#include<string>
#include<bitset>
#include<algorithm>
using namespace std;
typedef long long ll;
typedef long double ldb;
#define inf 99999999
#define pi acos(-1.0)
#define MOD 3
int n,m;
int data[4][805][805];

void solve()
{
    int i,j,k;
    for(i=0;i<n;i++){
        for(j=0;j<m;j++){
            data[3][i][j]=0;
        }
    }
    for(i=0;i<n;i++){
        for(k=0;k<m;k++){
            if(data[1][i][k]>0){
                for(j=0;j<m;j++){
                    data[3][i][j]=data[3][i][j]+data[1][i][k]*data[2][k][j];
                }
            }
        }
    }
}

int main()
{
    int i,j;
    while(scanf("%d",&n)!=EOF)
    {
        m=n;
        for(i=0;i<n;i++){
            for(j=0;j<n;j++){
                scanf("%d",&data[1][i][j]);
                data[1][i][j]%=3;
            }
        }

        for(i=0;i<n;i++){
            for(j=0;j<n;j++){
                scanf("%d",&data[2][i][j]);
                data[2][i][j]%=3;
            }
        }

        solve();
        for(i=0;i<n;i++){
            for(j=0;j<n;j++){
                if(j<n-1)printf("%d ",data[3][i][j]%3);
                else printf("%d\n",data[3][i][j]%3);
            }
        }
    }
    return 0;
}

代码二：用bitset

#include<iostream>
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<math.h>
#include<vector>
#include<map>
#include<set>
#include<queue>
#include<stack>
#include<string>
#include<bitset>
#include<algorithm>
using namespace std;
typedef long long ll;
typedef long double ldb;
#define inf 99999999
#define pi acos(-1.0)
#define MOD 3
#define maxn 805
bitset<1000>bt[maxn][3],ct[maxn][3];


int main()
{
    int n,m,i,j,c;
    while(scanf("%d",&n)!=EOF)
    {
        for(i=1;i<=n;i++){
            for(j=0;j<3;j++){
                bt[i][j].reset();
                ct[i][j].reset();
            }
        }
        for(i=1;i<=n;i++){
            for(j=1;j<=n;j++){
                scanf("%d",&c);
                bt[i][c%3].set(j);
            }
        }
        for(i=1;i<=n;i++){
            for(j=1;j<=n;j++){
                scanf("%d",&c);
                ct[j][c%3].set(i);
            }
        }
        for(i=1;i<=n;i++){
            for(j=1;j<=n;j++){
                c=((bt[i][1]&ct[j][1]).count()+(bt[i][1]&ct[j][2]).count()*2+(bt[i][2]&ct[j][1]).count()*2+(bt[i][2]&ct[j][2]).count() )%3;
                if(j==n)printf("%d\n",c);
                else printf("%d ",c);
            }
        }
    }
    return 0;
}