Matrix multiplication 2 (codevs3147)

Title Description Description
A matrix of two n*n is given, and M is queried for the values of the given sub-matrices in their product.

Enter a description input Description
The first row of two positive integers n,m.
The next n rows, each row n nonnegative integers, represent the first matrix.
The next n rows, each row n nonnegative integers, represent the second matrix.
Next M line, four positive integers per line a,b,c,d, asks for the product of the first matrix and the second matrix,
The element and in the sub-matrix of the vertex as a row b in column A and section C of Line D.

Outputs description Output Description
For each query, the output line is an integer that represents the answer to the query.
Sample input to sample
3 2
1 9 8
3 2 0
1 8 3
9 8 4
0 5 15
1 9 6
1 1 3 3
2 3 1 2

Sample output Sample Outputs
661
388
Data Size & Hint
"Data size and conventions"

For 30% of the data met, N <= 100.
For 100% of data satisfied, n <= 2000,m <= 50000, input data matrix elements < 100,a,b,
C,d <= N.
Exercises
Although the title is matrix multiplication, there is no relationship between the fast power and the matrix.
30% procedure:
Directly two matrix violence is multiplied, and then the violence is questioned. (PS: Training team's problem to give so much violence points)
100% procedure:
If you know enough about matrix multiplication, you can see that we can actually process each query in the complexity of O (n).
The matrix is a (X1,X2,Y1,Y2), the first matrix is a, the second matrix is B then the matrix A can be calculated by branch
Suppose the element of the first row (x1) of a matrix is p[i];
So according to the law of matrix multiplication
P[i] equals the elements of the X1 line of the A matrix and the elements of column I of the b matrix are multiplied, then added.
Sigma{p[i]} (y1<=i<=y2) is multiplied by the elements of line X1 of the A matrix, respectively, and the elements of the B matrix Y1 through the Y2 column, and then added.
Then we can see that this formula can be used to propose the element of the X1 line of the A matrix using the law of multiplication, then the sum of the resulting product is multiplied by the first and second elements of the B matrix.
The second line is the same as the third row,
We will find that each column in the B matrix can be extracted from the same
Then the sum of the matrices of the last query is multiplied by the elements of each row of the x1-x2 line of the A matrix and the elements and correspondence of each column of column y1-y2 of the B matrix.
So as long as the prefix of each row of the A matrix is preprocessed and
Each column of the B-matrix is prefixed and can be
Time complexity O (n^2+n*m);
Note Use scanf or read-in optimizations.
Code:

#include <iostream>#include <cstdio>using namespace Std;#define FOR (i,n) for (int i=1;i<=n;i++)#define READ (x) scanf ("%d", &x);int n,m,x,y,xx,yy,aa[2001][2001]={0},bb[2001][2001]={0},a,b;int Main () {read (n); read (m);For (i,n) for (j,n)    {read (a);Aa[i][j]=aa[i-1][j]+a;    }For (i,n) for (j,n)        {read (b);bb[i][j]=bb[i][j-1]+b;        }For (i,m)       {Read (x); Read (y); Read ( xx); Read (yy) ;long long ans=0;if (x>xx) swap (X,XX);if (y>yy) swap (Y,YY);For (j,n)ans+= (Long Long) (Aa[xx][j]-aa[x-1][j]) * (Long Long) (bb[j][yy]-bb[j][y-1]);printf ("%lld\n", ans);       }}

Matrix multiplication 2 (codevs3147)