Problem:
Score : 400 points

Problem Statement

We have a grid with H rows and W columns of squares.
Snuke is painting these squares in colors 1, 2, ..., N.
Here, the following conditions should be satisfied:

For each i (1 ≤ i ≤ N), there are exactly a_i squares painted in Color i. Here, a_1 + a_2 + ... + a_N = H W.

For each i (1 ≤ i ≤ N), the squares painted in Color i are 4-connected. That is, every square painted in Color i can be reached from every square painted in Color i by repeatedly traveling to a horizontally or vertically adjacent square painted in Color i.

Find a way to paint the squares so that the conditions are satisfied.
It can be shown that a solution always exists.

Constraints

1 ≤ H, W ≤ 100

1 ≤ N ≤ H W

a_i ≥ 1

a_1 + a_2 + ... + a_N = H W

Input

Input is given from Standard Input in the following format:

H W
N
a_1 a_2 ... a_N

Output

Print one way to paint the squares that satisfies the conditions.
Output in the following format:

c_{1 1} ... c_{1 W}
:
c_{H 1} ... c_{H W}

Here, c_{i j} is the color of the square at the i-th row from the top and j-th column from the left.

Sample Input 1

2 2
3
2 1 1

Sample Output 1

1 1
2 3

Below is an example of an invalid solution:

1 2
3 1

This is because the squares painted in Color 1 are not 4-connected.

Sample Input 2

3 5
5
1 2 3 4 5

Sample Output 2

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

Sample Input 3

1 1
1
1

Sample Output 3

1

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)


#include <algorithm>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <numeric>
#include <queue>
#include <vector>
using namespace std;
#define INF 1000000001;

int main() {
  int h, w, n;
  cin >> h >> w;
  cin >> n;
  int c[h * w];
  int num;
  int idx = 0;
  int color = 1;

  for (int i = 0; i < n; i++) {
    cin >> num;
    for (int j = 0; j < num; j++) {
      c[idx + j] = color;
    }
    idx += num;
    color++;
  }

  // for (int i = 0; i < h * w; i++) {
  //   cerr << c[i] << " ";
  // }
  // cerr << endl;

  for (int i = 0; i < h; i++) {
    if (i % 2 == 0) {
      for (int j = 0; j < w; j++) {  //そのまま
        cout << c[i * w + j] << " ";
      }
    } else {
      for (int j = w - 1; j >= 0; j--) {  //逆にする
        cout << c[i * w + j] << " ";
      }
    }
    cout << endl;
  }

  return 0;
}