Prime Ring Problem

                                                 Prime Ring Problem

                                              Time Limit: 2000ms                                         Memory Limit: 32768KB

                                                             64-bit integer IO format:      Java class name:

 

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.

 

Input

n (0 < n < 20).

 

Output

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. 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

简单的深度优先搜索题

源代码：

#include <iostream>
#include<cstring>
#include<cmath>
using namespace std;
int d[22];
int a[22];
int n,t=1;

int zs(int x)  //判断是否为质数
{
    for(int i=2;i<=(int)sqrt(x);i++)
        if(x%i==0)
           return 0;
        return 1;
}
void dfs(int x,int cnt)
{
    if(cnt==n)   //打表输出
    {
        for(int i=0;i<n-1;i++)
            cout<<a[i]<<" ";
        cout<<a[n-1]<<endl;
    }
    if(cnt>n)
        return;
    for(int i=2;i<=n;i++)
        if(!d[i]&&zs(i+x))
        {
            if(cnt==n-1)
                if(!zs(i+a[0]))  //不要忘记判断最后一个数和第一个数
                   continue;
            a[cnt]=i;
            d[i]=1;        //写入环中，标记为1
            dfs(i,cnt+1);
            d[i]=0;         //不符合要求，重新标记为0
        }
}
int main()
{
    while(cin>>n)
    {
        memset(d,0,sizeof(d));
        a[0]=1;
        d[1]=1;          //第一个必定为1，不要忘记标记
        cout<<"Case "<<t++<<":"<<endl;
        dfs(1,1);
        cout<<endl;
    }
    return 0;
}