"Codevs 3147" Matrix multiplication 2 2012 (WTF)

3147 Matrix multiplication 2 2012
Time limit: 1 s space limit: 64000 KB title level: Master Master
Title Description Description
gives two n*n matrices, M asks the number of the given sub-matrix in their product Values and.
* To prevent card evaluation, the data range has been reduced and the time limit is lowered. The
input describes the first line of input Description
with two positive integer n,m.
the next n rows, with n non-negative integers per row, represent the first matrix.
the next n rows, with n non-negative integers per row, represent the second matrix. The
Next M line, four positive integers per line a,b,c,d, indicates that the first matrix is queried for the product of the second matrix, and the element in the sub-matrix of row B and column C of line A is the vertex of the node. The
output describes output Description
for each query, outputting an integer line that represents the answer to the query.
Sample Input
3 2
1 9 8
3 2 0
1 8 3
9 8 4
0 5 All
1 9 6
1 1 3 3
2 3 1 2
Sample Output
661
388
data range and hints for the database size & Hint
"Data size and conventions"
for 40% data satisfaction, n <= 100,m <= 1000. The
satisfies 100% of the data, n <= 800,m <= 10000, the matrix element in the input data < 100,a,b,c,d <= N. The
data has gradients.

"This problem is very deceptive, I just began to obediently write the moment multiplication, however, because the simple matrix multiplication is already O (n^3), so add O (n^2) query on T, and then after the moment multiply to maintain the entire matrix prefix and, the query operation to the O (n), but still T"
"This problem is not actually a matrix multiplication, it can be pushed through the formula."
"By pushing the formula, you can see that this problem only need to calculate a, b two array prefix and (a array for each column prefix and b array for each row prefix and), and finally in the loop m, with the prefix and subtract, the interval is calculated and multiplied, so that the computation sub-matrix and the time of the calculation of O (n), The complexity of the core program segment is O (nm), and it can be over. 】
"Note: Be careful with MLE"

∑I=X1X2∑J=Y1Y2C (i,j) \sum_{i={x_1}}^{x_2}\sum_{j={y_1}}^{y_2} C (I,j)
=∑I=X1X2∑J=Y1Y2∑K=1NA[I][K]XB[K][J] =\sum_{i={x_1}}^{x_2}\sum_{j={y_1}}^{y_2}\sum_{k=1}^n A[i][k]xb[k][j]
=∑K=1N∑I=X1X2A[I][K]∑J=Y1Y2B[K][J] =\sum_{k=1}^n\sum_{i={x_1}}^{x_2}a[i][k] \sum_{j={y_1}}^{y_2} B[k][j]

 #include <cstdio> #include <cstring> #include <algorithm> using namespace
Std
int a[2001][2001],b[2001][2001];
int n,m,x1,y1,x2,y2;
    int main () {int i,j;
    scanf ("%d%d", &n,&m);
          for (I=1;i<=n;++i) for (j=1;j<=n;++j) {scanf ("%d", &a[i][j]);    
        A[I][J]+=A[I-1][J];
        } for (I=1;i<=n;++i) for (j=1;j<=n;++j) {scanf ("%d", &b[i][j]);
      B[I][J]+=B[I][J-1];
      } for (I=1;i<=m;++i) {scanf ("%d%d%d%d", &x1,&y1,&x2,&y2); 
      Long Long s1,s2;
      Long Long ans=0;
            for (j=1;j<=n;j++) {S1=a[max (x1,x2)][j]-a[min (x1,x2) -1][j];
            S2=b[j][max (y1,y2)]-b[j][min (Y1,y2)-1];
        ANS=ANS+S1*S2; 
    } printf ("%lld\n", ans);      
} return 0; }