Problem:
Score : 700 points

Problem Statement

We have a grid with H rows and W columns of squares.
We will represent the square at the i-th row from the top and j-th column from the left as (i,\ j).
Also, we will define the distance between the squares (i_1,\ j_1) and (i_2,\ j_2) as |i_1 - i_2| + |j_1 - j_2|.

Snuke is painting each square in red, yellow, green or blue.
Here, for a given positive integer d, he wants to satisfy the following condition:

No two squares with distance exactly d have the same color.

Find a way to paint the squares satisfying the condition.
It can be shown that a solution always exists.

Constraints

2 ≤ H, W ≤ 500

1 ≤ d ≤ H + W - 2

Input

Input is given from Standard Input in the following format:

H W d

Output

Print a way to paint the squares satisfying the condition, in the following format. If the square (i,\ j) is painted in red, yellow, green or blue, c_{ij} should be R, Y, G or B, respectively.

c_{11}c_{12}...c_{1W}
:
c_{H1}c_{H2}...c_{HW}

Sample Input 1

2 2 1

Sample Output 1

RY
GR

There are four pairs of squares with distance exactly 1.
As shown below, no two such squares have the same color.

(1,\ 1), (1,\ 2) : R, Y

(1,\ 2), (2,\ 2) : Y, R

(2,\ 2), (2,\ 1) : R, G

(2,\ 1), (1,\ 1) : G, R

Sample Input 2

2 3 2

Sample Output 2

RYB
RGB

There are six pairs of squares with distance exactly 2.
As shown below, no two such squares have the same color.

(1,\ 1) , (1,\ 3) : R , B

(1,\ 3) , (2,\ 2) : B , G

(2,\ 2) , (1,\ 1) : G , R

(2,\ 1) , (2,\ 3) : R , B

(2,\ 3) , (1,\ 2) : B , Y

(1,\ 2) , (2,\ 1) : Y , R

Solution in C (GCC 5.4.1) (judge verdict: Wrong Answer)

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

int tc[5];
void inittc(void) {
  int i;
  for (i = 0; i < 5; i++) {
    tc[i] = 0;
  }
}

int main() {
  int i, j;
  int h, w, d;
  int l, m;
  int t_i, t_j;

  int ans[503][503];
  char ans_c[501][501];
  for (i = 0; i < 503; i++) {
    for (j = 0; j < 503; j++) {
      ans[i][j] = 0;
    }
  }
  scanf("%d %d %d", &h, &w, &d);

  for (i = 0; i < h; i++) {
    for (j = 0; j < w; j++) {
      if (ans[i][j] != 0) {
        continue;
      }
      inittc();
      for (l = 0; l <= d;l++){
        t_i = i + l;
        t_j = j + (d - l);
        if(t_i >= 0 && t_i < h && t_j>=0 && t_j<w){
          tc[ans[t_i][t_j]]++;
        }
        t_i = i - l;
        t_j = j + (d - l);
        if(t_i >= 0 && t_i < h && t_j>=0 && t_j<w){
          tc[ans[t_i][t_j]]++;
        }
        t_i = i + l;
        t_j = j - (d - l);
        if(t_i >= 0 && t_i < h && t_j>=0 && t_j<w){
          tc[ans[t_i][t_j]]++;
        }
        t_i = i - l;
        t_j = j - (d - l);
        if(t_i >= 0 && t_i < h && t_j>=0 && t_j<w){
          tc[ans[t_i][t_j]]++;
        }
      }
      if(tc[1] == 0) {
        ans[i][j] = 1;
      }
      else if(tc[2] == 0) {
        ans[i][j] = 2;
      }
      else if(tc[3] == 0) {
        ans[i][j] = 3;
      }
      else if(tc[4] == 0) {
        ans[i][j] = 4;
      }
    }
  }

  for (i = 0; i < h;i++){
    for (j = 0; j < w;j++){
      if(ans[i][j] == 0){
        ans_c[i][j] = 'E';
      }
      else if(ans[i][j] == 1){
        ans_c[i][j] = 'R';
      }
      else if(ans[i][j] == 2){
        ans_c[i][j] = 'Y';
      }
      else if(ans[i][j] == 3){
        ans_c[i][j] = 'G';
      }
      else if(ans[i][j] == 4){
        ans_c[i][j] = 'B';
      }
      else{
        ans_c[i][j] = 'N';
      }
    }
    ans_c[i][j] = '\0';
  }

  for (i = 0; i < h;i++) {
    printf("%s\n", ans_c[i]);
  } return 0;
}