Prime Ring Problem HDU - 1016

A ring is compose of n circles as shown in diagram. Put natural number 1, 2, ..., n 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. 

 

Inputn (0 < n < 20). 
OutputThe 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. Print solutions in lexicographical order.

You are to write a program that completes above process. 

Print a blank line after each case. 
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
DFS 素数环
 1 #include <iostream>
 2 using namespace std;
 3 #include<string.h>
 4 #include<set>
 5 #include<stdio.h>
 6 #include<math.h>
 7 #include<queue>
 8 #include<map>
 9 #include<algorithm>
10 #include<cstdio>
11 #include<cmath>
12 #include<cstring>
13 #include <cstdio>
14 #include <cstdlib>
15 using namespace std;
16 
17 int a[25];
18 int TM[25];
19 int n;
20 
21 int QQQ(int x)
22 {
23     for(int i=2;i<=x/2;i++)
24     {
25         if(x%i==0)
26         return 0;
27     }
28     return 1;
29 }
30 
31 void dfs(int k,int CCC)
32 {
33     if(k==n)
34     {
35             if(QQQ(CCC+1))
36             {
37                 a[k]=CCC;
38                 int flag=1;
39                 for(int i=0;i<n;i++)
40                 {
41                     if(flag)
42                     {
43                         flag=0;
44                         cout<<a[i];
45                         continue;
46                     }
47                     cout<<' '<<a[i];
48                 }
49                 cout<<endl;
50             }
51             return;
52     }
53     for(int i=2;i<=n;i++)
54     {
55             if(!TM[i]&&QQQ(CCC+i))
56             {
57             TM[i]=1;
58             a[k]=i;
59             dfs(k+1,i);
60             TM[i]=0;
61             }
62     }
63 }
64 
65 int main()
66 {
67     int add=0;
68     while(cin>>n)
69     {
70         cout<<"Case "<<++add<<":"<<endl;
71     memset(TM,0,sizeof(TM));
72     a[0]=1;
73     dfs(1,1);
74     cout<<endl;
75     }
76     return 0;
77 }
View Code

 

posted @ 2017-08-02 10:54  小小超plus  阅读(168)  评论(0)  编辑  收藏  举报