Problem F Necklace
Time Limit: 2 Seconds Memory Limit: 65536 KB
A necklace consists of several beads which are forming a closed loop. Each bead can only be one of the N distinct colors. First, we define concept "pair-of-colors" which contains two different colors, like (blue, red) or (white, black). Let's consider a "pair-of-colors" set S which contains all possible color pairs. For example, when there are 3 distinct colors: blue, red, white; S will be { (blue, red), (blue, white), (red, blue), (red, white), (white, red), (white, blue) }.
Now, let's add the colors of each two adjacent beads on a necklace as "pair-of-colors" to an empty set. If the result set is the same with S on all distinct colors appeared in the necklace, the necklace is perfect. Cissy wants to know what the shortest perfect necklace is when there are infinite beads with all N distinct colors.
Input
The input consists of several test cases. For each case, there are a positive integers N indicating the number of different bead colors.
N is odd integer, in the range of [3,49].
Proceed to the end of file.
Output
For each test case, output the minimal length of the perfect necklace in one line followed with the configuration of necklace in the second line. Here use integers in [0, N-1] to indicate the color of the bead.
If there are more than one configuration with the minimal length, output any one of them.
Sample Input
3
7
Sample Output
3
2 1 0
21
4 0 5 1 6 2 3 4 2 5 3 6 4 5 6 0 1 2 0 3 1
题解：
数学组合、DFS。对0~N-1进行两两组合，由于是一条链上，使用深度优先。
如3：0->1->2->0
5：0->1->2->0->3->1->4->2->3->4->0
代码实现：
#include <iostream>
#include <cstdio>
using namespace std;
const int maxn = 100;
const int maxd = 1e7;
bool visited[maxn][maxn];
int d[maxd];
int init(){
for(int i=0;i<maxn;i++){
for(int j = 0;j<maxn;j++){
visited[i][j] = false;
}
}
return 0;
}
int t;
int dfs(int v,int n){
for(int i=0;i<n;i++){
if(i!=v && !visited[i][v]){
visited[i][v] = visited[v][i] = true;
dfs(i,n);
}
}
d[t] = v;
t++;
return 0;
}
int main()
{
int n;
while(~scanf("%d",&n)){
init();
t = 0;
dfs(0,n);
printf("%d\n",t-1);
printf("%d",d[1]);
for(int i=2;i<t;i++){
printf(" %d",d[i]);
}
printf("\n");
}
return 0;
}