hdu 4920 Matrix multiplication bitset优化常数

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+,M=1e6+,inf=1e9+;
const ll INF=1e18+,mod=;
int n;
bitset<N>a[N][],b[N][];
int ans(int i,int j)
{
    int u=(a[i][]&b[j][]).count(),v=(a[i][]&b[j][]).count();
    int x=(a[i][]&b[j][]).count(),y=(b[j][]&a[i][]).count();
    return u+v*+x*+y*;
}
int main()
{
    while(~scanf("%d",&n))
    {
        for(int i=;i<n;i++)
            for(int j=;j<=;j++)
            a[i][j].reset(),b[i][j].reset();
        for(int i=; i<n; i++)
            for(int j=; j<n; j++)
            {
                int x;
                scanf("%d",&x);
                a[i][x%].set(j);
            }
        for(int i=; i<n; i++)
            for(int j=; j<n; j++)
            {
                int x;
                scanf("%d",&x);
                b[j][x%].set(i);
            }
        for(int i=;i<n;i++)
        {
            for(int j=;j<n;j++)
                printf("%d%c",ans(i,j)%,(j!=n-?' ':'\n'));
        }
    }
    return ;
}