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 k 3 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.
The first line contains integer k (1 ≤ k ≤ 100) , which is size of the cube.
Print -1 if there is no solution. Otherwise, print the required painting of the cube consequently by layers. Print a k × k matrix in the first k lines, showing how the first layer of the cube should be painted. In the following k lines print a k × k matrix — the way the second layer should be painted. And so on to the last k -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.
1
-1
2
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;
}
