Matrix multiplication（bitset）

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Matrix multiplication

Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 1928 Accepted Submission(s): 872

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

10120 12 34 56 7

Sample Output

00 12 1

Author

Xiaoxu Guo (ftiasch)

Source

2014 Multi-University Training Contest 5

第一次接触bitset，纠结了好久，也光是调试就调试了二十多分钟，听说还可以用n3解决，不过要改变循环的顺序；
下面这句是抄大神的；

这样写会超时：            for (int i=1; i<=n; i++)                for (int j=1; j<=n; j++)                    for (int k=1; k<=n; k++)                        c[i][j]+=a[i][k]*b[k][j];这样写就能过:        for (int k=1; k<=n; k++)            for (int i=1; i<=n; i++)                for (int j=1; j<=n; j++)                    c[i][j]+=a[i][k]*b[k][j];

我们知道内存中二维数组是以行为单位连续存储的，逐列访问将会每次跳1000*4(bytes)。根据cpu cache的替换策略，将会有大量的cache失效。时间居然会相差很多。

以下是我的bitset代码：

AC 代码

/****************************************************** / /*                                                   /*  *   ***********                                      *  *           *      Auther:     ZSGG                  *  *         *                                          *  *       *          Name:        bitset               *  *     *                                              *  *   ***********    Algorithm:                        *  *                                                    *  *//*******************************************************/#include <iostream>#include <cstring>#include <cstdio>#include <cmath>#include <bitset>#include <queue>#include <vector>#include <cstdlib>#include <algorithm>#define ls u << 1#define rs u << 1 | 1#define lson l, mid, u << 1#define rson mid + 1, r, u << 1 | 1using namespace std;const int M = 810;int b[M][M],n;int main(){//    freopen("in.txt","r",stdin);//    freopen("out.txt","w",stdout);    bitset<810>t[M][3]; //Every elements has eight hundrys and ten elements;    bitset<810>s[M][3];    while(~scanf("%d",&n))    {        for(int i = 1; i <= n; i++) //清空bitset容器；        {            for(int j = 1; j < 3; j++)             {                t[i][j].reset();                 s[i][j].reset();            }        }        for(int i = 1; i <= n; i++)        {            for(int j = 1; j <= n; j++)            {                int x;                scanf("%d",&x);                x %= 3;                if(x) t[i][x][j] = 1;  //在t[i][x]这个元素的第j个位置设置为1；            }        }                // 用bitset可以固定位置，这可以提高效率，使内层循环从n缩小为2 * 2； 详细看下面的代码；        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++)        {            for(int j = 1; j <= n; j++)            {                if(b[j][i]) s[i][b[j][i]][j] = 1; //按列进bitset            }        }               /*核心代码*/        /*************************************************************/        for(int i = 1; i <= n; i++)        {            for(int j = 1; j <= n; j++)            {                   int res = 0;                   for(int p = 1; p <= 2; p++)                    {                       for(int k = 1; k <= 2; k++)                       {                           res += (t[i][p] & s[j][k]).count() * p * k; //  求出行和列的并集个数；如 0010 1010 交集个数为1；                       }                   }                   printf(j == n ? "%d\n" : "%d ", res % 3);            }        }                /************************************************************/    }    return 0;}AC 代码：

/****************************************************** / /*                                                   /*  *   ***********                                      *  *           *      Auther:     ZSGG                  *  *         *                                          *  *       *          Name:        bitset               *  *     *                                              *  *   ***********    Algorithm:                        *  *                                                    *  *//*******************************************************/#include <iostream>#include <cstring>#include <cstdio>#include <cmath>#include <bitset>#include <queue>#include <vector>#include <cstdlib>#include <algorithm>#define ls u << 1#define rs u << 1 | 1#define lson l, mid, u << 1#define rson mid + 1, r, u << 1 | 1using namespace std;const int M = 810;int a[M][M],b[M][M],c[M][M];int main(){//    freopen("in.txt","r",stdin);//    freopen("out.txt","w",stdout);    int n;    while(~scanf("%d",&n))    {        memset(c,0,sizeof(c));        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 k = 1; k <= n; k++)        {            for(int i = 1; i <= n; i++)            {                for(int j = 1; j <= n; j++)                {                    c[i][j] += b[k][j] * a[i][k];                }            }        }        for(int i = 1; i <= n; i++)        {            for(int j  = 1; j <= n; j++)            {                if(j == n) printf("%d\n",c[i][j] % 3);                else printf("%d ",c[i][j] % 3);            }        }    }    return 0;}

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