Prime Ring Problem HDU

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

#include <iostream>
using namespace std;
#include<string.h>
#include<set>
#include<stdio.h>
#include<math.h>
#include<queue>
#include<map>
#include<algorithm>
#include<cstdio>
#include<cmath>
#include<cstring>
#include <cstdio>
#include <cstdlib>
using namespace std;

int a[25];
int TM[25];
int n;

int QQQ(int x)
{
    for(int i=2;i<=x/2;i++)
    {
        if(x%i==0)
        return 0;
    }
    return 1;
}

void dfs(int k,int CCC)
{
    if(k==n)
    {
            if(QQQ(CCC+1))
            {
                a[k]=CCC;
                int flag=1;
                for(int i=0;i<n;i++)
                {
                    if(flag)
                    {
                        flag=0;
                        cout<<a[i];
                        continue;
                    }
                    cout<<' '<<a[i];
                }
                cout<<endl;
            }
            return;
    }
    for(int i=2;i<=n;i++)
    {
            if(!TM[i]&&QQQ(CCC+i))
            {
            TM[i]=1;
            a[k]=i;
            dfs(k+1,i);
            TM[i]=0;
            }
    }
}

int main()
{
    int add=0;
    while(cin>>n)
    {
        cout<<"Case "<<++add<<":"<<endl;
    memset(TM,0,sizeof(TM));
    a[0]=1;
    dfs(1,1);
    cout<<endl;
    }
    return 0;
}