HDU 1016：Prime Ring Problem

Prime Ring Problem

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 25156    Accepted Submission(s): 11234

Problem Description

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

</pre><pre name="code" class="cpp">#include<cstdio>
#include<stdlib.h>
#include<cstring>
#include<algorithm>
#include<iostream>
 
using namespace std;
 
int date[25]= {1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20};
int ans[25];
int n, cnt;
int flag;
 
void dfs(int step)
{
    int temp;
    if(step==n)       //深搜究竟时注意推断首尾是否符合要求
    {
        flag=1;
        temp = ans[0]+ans[n-1];
        if(flag)
        {
            for(int i=2; i*i<=temp; i++)
                if(temp%i==0)
                {
                    flag=0;
                    break;
                }
        }
        if(flag)
        {
            printf("%d", ans[0]);
            for(int i=1; i<n; i++)
                printf(" %d", ans[i]);
            printf("\n");
        }
    }
    else
    {
        for(int i=1; i<n; i++)
        {
            if(date[i])
            {
                temp=ans[cnt-1]+date[i];
                flag=1;
                for(int j=2; j*j<=temp; j++)
                {
                    if(temp%j==0)
                    {
                        flag=0;
                        break;
                    }
                }
                if(flag)
                {
                    ans[cnt++]=date[i];
                    date[i]=0;
                    dfs(step+1);
                    date[i]=ans[--cnt];   //记得将数据的还原处理
                    ans[cnt]=0;
                }
            }
        }
    }
}
 
int main()
{
    int count=1;
    while(scanf("%d", &n)!=EOF)
    {
        memset(ans, 0, sizeof(ans));
        cnt=1;
        ans[0]=1;
        printf("Case %d:\n",count++);
        dfs(1);            //从1開始进入搜索
        printf("\n");      //每次输出后记得加个换行
    }
 
    return 0;
}