Matrix multiplication hdu4920

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Problem DescriptionGiven 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. 

 

InputThe 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 Aij. The next n lines describe the matrix B in similar format (0≤Aij,Bij≤109). 

 

OutputFor each tests:

Print n lines. Each of them contain n integers -- the matrix A×B in similar format. 

 

Sample Input10120 12 34 56 7 

 

Sample Output0 0 1 2 11，忽視0  去做。

 1 #include"stdio.h" 2 #include"string.h" 3 int a[801][801],b[801][801]; 4 int a1[801][801],b1[801][801]; 5 int c[801][801]; 6 int main() 7 { 8     int n,i,j,k; 9     while(scanf("%d",&n)==1)10     {11         memset(a,0,sizeof(a));12         memset(b,0,sizeof(b));13         memset(c,0,sizeof(c));14         memset(a1,0,sizeof(a1));15         memset(b1,0,sizeof(b1));16         for(i=1;i<=n;i++)17             for(j=1;j<=n;j++)18             {19                 scanf("%d",&a[i][j]);20                 a[i][j]%=3;21             }22         for(i=1;i<=n;i++)23             for(j=1;j<=n;j++)24             {25                 scanf("%d",&b[i][j]);26                 b[i][j]%=3;27             }28         for(i=1;i<=n;i++)29         {30             int pre=-1;31             for(j=n;j>=0;j--)32             {33                 a1[i][j]=pre;34                 if(a[i][j])35                     pre=j;36             }37         }38         for(i=1;i<=n;i++)39         {40             int pre=-1;41             for(j=n;j>=0;j--)42             {43                 b1[i][j]=pre;44                 if(b[i][j])45                     pre=j;46             }47         }48         for(i=1;i<=n;i++)49             for(j=a1[i][0];j+1;j=a1[i][j])50                 for(k=b1[j][0];k+1;k=b1[j][k])51                     c[i][k]+=a[i][j]*b[j][k];52         for(i=1;i<=n;i++)53         {54             for(j=1;j<n;j++)55                 printf("%d ",c[i][j]%3);56             printf("%d\n",c[i][j]%3);57         }58     }59     return 0;60 }

 

我們知道記憶體中二維數組是以行為單位連續儲存的，逐列訪問將會每次跳1000*4(bytes)。根據cpu cache的替換策略，將會有大量的cache失效。

時間居然會相差很多。 可見利用好cpu cache最佳化我們的程式，是非常有必要掌握的技能。
平時寫程式時，也應當盡量使cpu對記憶體的訪問，是儘可能連續的

/*    Name: Matrix multiplication    Copyright: Shangli Cloud    Author: Shangli Cloud    Date: 05/08/14 20:46    Description: 轉置 *//*#include"iostream"#include"cstdio"#include"cstring"using namespace std;const int ms=801;const int mod=3;*/#include"stdio.h"#include"string.h"//int a[ms][ms],b[ms][ms],c[ms][ms];#define mod 3int a[801][801],b[801][801],c[801][801];int main(){    int n,x,i,j,k;    while(scanf("%d",&n)==1)    {        for(i=1;i<=n;i++)            for(j=1;j<=n;j++)            {                scanf("%d",&x);                a[i][j]=x%mod;            }        for(i=1;i<=n;i++)            for(j=1;j<=n;j++)            {                scanf("%d",&x);                b[j][i]=x%mod;            }        for(i=1;i<=n;i++)            for(j=1;j<=n;j++)            {                c[i][j]=0;                for(k=1;k<=n;k++)                {                    //c[i][j]+=a[i][k]*b[j][k]%mod;多了個mod就逾時，                     c[i][j]+=a[i][k]*b[j][k];//1656ms,多個Mod就超過2s.                 }            if(j<n)                printf("%d ",c[i][j]%mod);            else                printf("%d\n",c[i][j]%mod);            }    }    return 0;}