Matrix multiplication 2 (Codevs no.3147)

2016-06-01 17:33:30

Title Link: Matrix multiplication 2 (Codevs no.3147)

Main topic:

Given two square matrices of the same size, a, B. Multiple queries, each time a matrix of a sub-matrix of all elements of the and.

Solution:

First think of violence.

Preprocessing n^3, asking for a mock sweep, this constant is just an instant explosion.

Of course, it can be optimized.

Listing the element expressions of the sub-matrices, you will find some elements//matrix multiplication 2 (Codevs no.3147)
Matrix multiplication
#include <stdio.h>
#include <algorithm>
using namespace Std;
const int maxn=2010;
int MAP1[MAXN][MAXN];
int MAP2[MAXN][MAXN];
Long long ans;
int n,m;
int a,b,c,d;
int main ()
{
scanf ("%d%d", &n,&m);
for (int i=1;i<=n;i++)
{
for (int j=1;j<=n;j++)
{
scanf ("%d", &map1[i][j]);
MAP1[I][J]+=MAP1[I-1][J];
}
}
for (int i=1;i<=n;i++)
{
for (int j=1;j<=n;j++)
{
scanf ("%d", &map2[i][j]);
MAP2[I][J]+=MAP2[I][J-1];
}
}
for (int i=1;i<=m;i++)
{
ans=0;
scanf ("%d%d%d%d", &a,&b,&c,&d);
for (int i=1;i<=n;i++)
{
int Sumxmap2=map2[i][max (B,D)]-map2[i][min (b,d)-1];
int Sumymap1=map1[max (A,C)][i]-map1[min (a,c) -1][i];
ans+= (long Long) sumxmap2*sumymap1;
}
printf ("%lld\n", ans);
}
return 0;
The product of the} can be used to multiply the distribution law

So we simply give up the exact value of the sub-matrix and combine the data according to the characteristics of the distributive law.

We prefix the column with a (multiply by line) and, for B (multiplication by the time), prefix the line and

Finally, all the elements can be represented as Sigma ((A[i][dbound]-a[i][ubound-1]) * (b[i][rbound]-b[i][lbound-1]);

(Ubound,dbound,lbound,rbound represents the lower and upper sides of the small matrix)

1 //Matrix multiplication 2 (Codevs no.3147)2 //matrix multiplication3#include <stdio.h>4#include <algorithm>5 using namespacestd;6 Const intmaxn= .;7 intMAP1[MAXN][MAXN];8 intMAP2[MAXN][MAXN];9 Long Longans;Ten intn,m; One inta,b,c,d; A intMain () - { -scanf"%d%d",&n,&M); the      for(intI=1; i<=n;i++) -     { -          for(intj=1; j<=n;j++) -         { +scanf"%d",&map1[i][j]); -map1[i][j]+=map1[i-1][j]; +         } A     } at      for(intI=1; i<=n;i++) -     { -          for(intj=1; j<=n;j++) -         { -scanf"%d",&map2[i][j]); -map2[i][j]+=map2[i][j-1]; in         } -     } to      for(intI=1; i<=m;i++) +     { -ans=0; thescanf"%d %d%d%d",&a,&b,&c,&d); *          for(intI=1; i<=n;i++) $         {Panax Notoginseng             intSumxmap2=map2[i][max (b,d)]-map2[i][min (b,d)-1]; -             intSumymap1=map1[max (a,c)][i]-map1[min (a,c)-1][i]; theans+= (Long Long) sumxmap2*Sumymap1; +         } Aprintf"%lld\n", ans); the     } +     return 0; -}

Matrix multiplication 2 (Codevs no.3147)