hdu 4920 Matrix multiplication bitset优化常数

Posted 2020-08-26 xjhz

Matrix multiplication

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)

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

题意：给你两个矩阵，求矩阵相乘%3；

思路：正常思路n*n*n求解（运气好可以ac。。。），看了下数据范围由于mod=3，所以利用bitset标记每行1，2，0的位置，取并集就可以得到相同位置的个数；

　　　这样就可以优化最后的一个n。详见代码；

#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define pi (4*atan(1.0))
#define eps 1e-14
const int N=1e3+10,M=1e6+10,inf=1e9+10;
const ll INF=1e18+10,mod=2147493647;
int n;
bitset<N>a[N][5],b[N][5];
int ans(int i,int j)
{
    int u=(a[i][1]&b[j][1]).count(),v=(a[i][1]&b[j][2]).count();
    int x=(a[i][2]&b[j][1]).count(),y=(b[j][2]&a[i][2]).count();
    return u+v*2+x*2+y*4;
}
int main()
{
    while(~scanf("%d",&n))
    {
        for(int i=0;i<n;i++)
            for(int j=0;j<=3;j++)
            a[i][j].reset(),b[i][j].reset();
        for(int i=0; i<n; i++)
            for(int j=0; j<n; j++)
            {
                int x;
                scanf("%d",&x);
                a[i][x%3].set(j);
            }
        for(int i=0; i<n; i++)
            for(int j=0; j<n; j++)
            {
                int x;
                scanf("%d",&x);
                b[j][x%3].set(i);
            }
        for(int i=0;i<n;i++)
        {
            for(int j=0;j<n;j++)
                printf("%d%c",ans(i,j)%3,(j!=n-1?‘ ‘:‘\n‘));
        }
    }
    return 0;
}