1 second
256 megabytes
standard input
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.
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 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.
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;
}