cf 323A A. Black-and-White Cube 立体构造

A. Black-and-White Cube

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

You are given a cube of size k × k × k, which consists of unit cubes. Two unit cubes are considered neighbouring, if they have common face.

Your task is to paint each of k³ unit cubes one of two colours (black or white), so that the following conditions must be satisfied:

• each white cube has exactly 2 neighbouring cubes of white color;
• each black cube has exactly 2 neighbouring cubes of black color.

Input

The first line contains integer k (1 ≤ k ≤ 100), which is size of the cube.

Output

Print -1 if there is no solution. Otherwise, print the required painting of the cube consequently by layers. Print ak × k matrix in the first k lines, showing how the first layer of the cube should be painted. In the followingk lines print a k × k matrix — the way the second layer should be painted. And so on to the lastk-th layer. Note that orientation of the cube in the space does not matter.

Mark a white unit cube with symbol "w" and a black one with "b". Use the format of output data, given in the test samples. You may print extra empty lines, they will be ignored.

Sample test(s)

Input

1

Output

-1

Input

2

Output

bb
ww

bb
ww

题目意思，输出一个k*k*k的立体模型，使每个b的旁边有2个b，每个k的旁边有2个k
题解：

题目说的输出描述有点问题，它说的是先输出你的第1层，再输出k层从1到k的你的模型。实际上只用输出你构造的模型的1-k层。

这里提供两种构造方法

第1种

bbwwbb

bbwwbb

wwbbww

wwbbww

bbwwbb

bbwwbb

这种就是4个4个的。。以后每层就把上一层取反就OK了

还有一种是

bbbbbb

bwwwwb

bwbbwb

bwbbwb

bwwwwb

bbbbbb

类似这种不断将最外层围起来的构造方式，同理，以后的每层就把上一层取反就OK了。。

构造方法不唯一的。

至于 k 为奇数时无解我无法证明这个。。。

/*
 * @author ipqhjjybj
 * @date  20130709
 *
 */
#include <cstdio>
#include <cstdlib>

int main(){
    int k;
    scanf("%d",&k);
    if(k&1) {
        puts("-1");
        return 0;
    }
    for(int i=0;i<k;i++){
        for(int j=0;j<k;j++){
            for(int z=0;z<k;z++){
               putchar(((j>>1)&1)^((z>>1)&1)^(i&1)?'w':'b');
            }
            putchar('\n');
        }
        putchar('\n');
    }
    return 0;
}