A - Prime Ring Problem

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

题目分析 ：

　　　　1. 1的位置不变，从1开始DFS深搜；　　　　2. 素数函数的效率决定了代码是否超时；

错误点 ：   　　1. 素数的判断有误；　　　　2. while（true） 的形式有误，无法跳出，时间超时；

#include <iostream>
#include <math.h>
#include <vector>
#include <cstring>
using namespace std;
int n,m,array[25],mark[25],step,k=0;

bool Exame_prime(int x,int y)
{
    int sum = x+y;
    for(int i=2;i*i<=sum;i++)
        if(sum%i==0)
            return true;
    return false;
}

void DFS(int x)
{
    step++;
    array[step] = x;
    if(step==n)
    {
        if(!Exame_prime(array[step],1))
        {
            for(int i=1;i<=step;i++)
            {
                if(i!=step)
                    cout << array[i] << " ";
                else
                    cout << array[i];
            }
            cout << endl;
        }
        return;
    }

    for(int i=2;i<=n;i++)
    {
        //cout <<"|||| :" <<  x <<"  " <<i << "  " <<mark[i] <<"  "  << ++k <<  "\n";
        if(!Exame_prime(x,i) && !mark[i])
        {
            mark[i]=1;
            DFS(i);
            mark[i]=0;
            step--;//不可省略，step为常数，不会回溯；
        }
    }
}

int main()
{
    int m=0;
    while(cin >> n)
    {
        memset(mark,0,sizeof(mark));
        step=0;
        cout << "Case " << ++m <<":" << "\n";
        DFS(1);
        cout << endl;
    }
    return 0;
}