素数环

Description

A ring is composed of n (even number) circles as shown in diagram. Put natural numbers into each circle separately, and the sum of numbers in two adjacent circles should be a prime.

Note: the number of first circle should always be 1.

The output format is shown as sample below. Each row represents a series of circle numbers in the ring beginning from 1 clockwisely and anticlockwisely. The order of numbers must satisfy the above requirements.

You are to write a program that completes above process.

Sample Input

6
8

Sample Output

Case 1:
1 4 3 2 5 6
1 6 5 2 3 4

Case 2:
1 2 3 8 5 6 7 4
1 2 5 8 3 4 7 6
1 4 7 6 5 8 3 2
1 6 7 4 3 8 5 2

因为起点已经确定为1；

只要用dfs回溯就行；

同是要注意使用vis[]数组判重；

 1 #include<cstdio>
 2 #include<cstring>
 3 #include<algorithm>
 4 using namespace std;
 5 
 6 int is_prime(int x) {
 7   for(int i = 2; i*i <= x; i++)
 8     if(x % i == 0) return 0;
 9   return 1;
10 }
11 
12 int n, A[50];
13 bool isp[50],vis[50];
14 void dfs(int cur) {
15   if(cur == n && isp[A[0]+A[n-1]]){
16     for(int i = 0; i < n; i++){
17       if(i != 0) printf(" ");
18       printf("%d", A[i]);
19     }
20     printf("
");
21   }
22   else for(int i = 2; i <= n; i++)
23     if(!vis[i] && isp[i+A[cur-1]]){
24       A[cur] = i;
25       vis[i] = 1;
26       dfs(cur+1);
27       vis[i] = 0;
28     }
29 }
30 
31 int main() {
32   int kase = 0;
33   while(scanf("%d", &n) == 1 && n > 0){
34     if(kase > 0) printf("
");
35     printf("Case %d:
", ++kase);
36     for(int i = 2; i <= n*2; i++) isp[i] = is_prime(i);
37     memset(vis, 0, sizeof(vis));
38     A[0] = 1;
39     dfs(1);
40   }
41   return 0;
42 }

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