HDU4920：Matrix multiplication（思维 & bitset）

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Matrix multiplication

Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 4792 Accepted Submission(s): 1855

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

题意：输出两个n阶矩阵相乘的结果。

思路：这题主要是解决超时的问题，由于结果要模3，如果数据是随机的话，有1/3的数字会是0，那么更改一下循环嵌套的顺序，遇到0的直接continue就不会超时了。实际上，更改了循环顺序后，无需判断0都能AC，引用 Shangli Cloud 的话：

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];为什么？----------------------------------------------------------------------------------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。 可见利用好cpu cache优化我们的程序，是非常有必要掌握的技能。平时写程序时，也应当尽量使cpu对内存的访问，是尽可能连续的。

# include <iostream># include <cstdio>using namespace std;int a[801][801], b[801][801], c[801][801];int main(){    int n;    while(~scanf("%d",&n))    {        for(int i=0; i<n; ++i)            for(int j=0; j<n; ++j)                scanf("%d",&a[i][j]), a[i][j]%=3, c[i][j]=0;        for(int i=0; i<n; ++i)            for(int j=0; j<n; ++j)                scanf("%d",&b[i][j]),b[i][j]%=3;        for(int k=0; k<n; ++k)            for(int i=0; i<n; ++i)                for(int j=0; j<n; ++j)                    if(a[i][k])//因此这个注释掉也不会超时。                        c[i][j] += a[i][k]*b[k][j];        for(int i=0; i<n; ++i)            for(int j=0; j<n; ++j)                printf("%d%c",c[i][j]%3,j==n-1?'\n':' ');    }    return 0;}

下面是用Bitset容器的做法。

# include <iostream># include <cstdio># include <bitset>using namespace std;bitset<803>a[803][4], b[803][4];int fun(int i, int j){    int x1 = (a[i][1]&b[j][1]).count();    int x2 = (a[i][1]&b[j][2]).count();    int x3 = (a[i][2]&b[j][1]).count();    int x4 = (a[i][2]&b[j][2]).count();    return x1 + ((x2+x3)<<1) + (x4<<2);}int main(){    int n, t;    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)            {                scanf("%d",&t);                a[i][t%3].set(j);            }        for(int i=0; i<n; ++i)            for(int j=0; j<n; ++j)            {                scanf("%d",&t);                b[j][t%3].set(i);            }        for(int i=0; i<n; ++i)            for(int j=0; j<n; ++j)                printf("%d%c",fun(i, j)%3,j==n-1?'\n':' ');    }    return 0;}

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