Problem:
Score : 400 points

Problem Statement

We have a grid with H rows and W columns. At first, all cells were painted white.

Snuke painted N of these cells. The i-th ( 1 \leq i \leq N ) cell he painted is the cell at the a_i-th row and b_i-th column.

Compute the following:

For each integer j ( 0 \leq j \leq 9 ), how many subrectangles of size 3×3 of the grid contains exactly j black cells, after Snuke painted N cells?

Constraints

3 \leq H \leq 10^9

3 \leq W \leq 10^9

0 \leq N \leq min(10^5,H×W)

1 \leq a_i \leq H (1 \leq i \leq N)

1 \leq b_i \leq W (1 \leq i \leq N)

(a_i, b_i) \neq (a_j, b_j) (i \neq j)

Input

The input is given from Standard Input in the following format:

H W N
a_1 b_1
:
a_N b_N

Output

Print 10 lines.
The (j+1)-th ( 0 \leq j \leq 9 ) line should contain the number of the subrectangles of size 3×3 of the grid that contains exactly j black cells.

Sample Input 1

4 5 8
1 1
1 4
1 5
2 3
3 1
3 2
3 4
4 4

Sample Output 1

0
0
0
2
4
0
0
0
0
0

There are six subrectangles of size 3×3. Two of them contain three black cells each, and the remaining four contain four black cells each.

Sample Input 2

10 10 20
1 1
1 4
1 9
2 5
3 10
4 2
4 7
5 9
6 4
6 6
6 7
7 1
7 3
7 7
8 1
8 5
8 10
9 2
10 4
10 9

Sample Output 2

4
26
22
10
2
0
0
0
0
0

Sample Input 3

1000000000 1000000000 0

Sample Output 3

999999996000000004
0
0
0
0
0
0
0
0
0

Solution in C (GCC 5.4.1) (judge verdict: Time Limit Exceeded)

#include <stdio.h>
#include <stdlib.h>
#include <math.h>

#define MAX_N   100000

int mod(int a, int b){
    return (a % b >= 0)?  a % b : a % b + b;
}

int main(){
    int m, n, i, j, H, W, N, hashnum=(int)(MAX_N*9*1.5),
        a[MAX_N], b[MAX_N], nonzero=0, result[10]={},
        bignum[20], tmp[20], carry=0, h[10], w[10], order;
    
    scanf("%d %d %d", &H, &W, &N);
    for (n=0; n<N; n++) scanf("%*c%d %d", &a[n], &b[n]);
    
    hashnum = (int)(N*9*1.5);
    
    struct blockmodel{
        int count, center[2];
    } *block;
    block = calloc(hashnum, sizeof(struct blockmodel));
    
    for (n=0; n<N; n++){
        for (i=-1; i<=1; i++){
            if (a[n]-1+i<1 || H-2<a[n]-1+i) continue;
            for (j=-1; j<=1; j++){
                if (b[n]-1+j<1 || W-2<b[n]-1+j) continue;
                m = ((a[n]+b[n]-2+i+j)*17) % hashnum;
                while (block[m].count != 0) {
                    if (block[m].center[0] == a[n]-1+i
                        && block[m].center[1] == b[n]-1+j) break;
                    m = (m+1) % hashnum;
                }
                block[m].center[0] = a[n]-1+i;
                block[m].center[1] = b[n]-1+j;
                block[m].count++;
            }
        }
    }
    
    for (m=0; m<hashnum; m++){
        if (block[m].count != 0){
            result[block[m].count]++;
            nonzero++;
        }
    }
    
    for (i=0; i<10; i++){
        h[i] = ((H-2) % (int)pow(10, i+1))/(int)pow(10, i);
        w[i] = ((W-2) % (int)pow(10, i+1))/(int)pow(10, i);
    }
    for (i=0; i<10; i++){
        for (j=0; j<10; j++){
            bignum[i+j] += (h[i]*w[j]+carry) % 10;
            carry = (h[i]*w[j]+carry)/10;
        }
    }
    carry = 0;
    for (i=0; i<20; i++){
        tmp[i] = (bignum[i] + carry) % 10;
        carry = (bignum[i]+carry)/10;
    }
    carry = 0;
    tmp[0] = tmp[0] - nonzero;
    for (i=0; i<20; i++){
        bignum[i] = mod(tmp[i]+carry, 10);
        carry = (tmp[i]+carry-9)/10;
    }
    
    for (order=19; order>=1; order--){
        if (bignum[order] != 0) break;
    }
    
    for (i=order; i>=0; i--){
        printf("%d", bignum[i]);
    }
    printf("\n");
    
    for (i=1; i<10; i++){
        printf("%d\n", result[i]);
    }

    free(block);
    
    return 0;
}