Prime Ring Problem

Time Limit: 400

Prime Ring Problem

0/2000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 28085 Accepted Submission(s): 12501

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

源码：

#include <stdio.h>
#include <string.h>
#include <stdlib.h>

int n;
int prime[30]={1,1,0};
int way[30];    //存储最后结果序列
int visit[30];  //vis表示是否訪问
void isprime() //不是素数标记为1 
{
  int i,j;
  for(i=2;i<30;i++)
  {
    for(j=2*i;j<30;j+=i) 
    {
      prime[j]=1;
    }
  }
/*
  for(i = 0;i < 30;i ++)
    printf("prime[%d]=%d\n",i,prime[i]);
*/
}


/*
素数序列的开头是这种：
2，3，5，7。11，13。17，19。23，29，
31，37，41。43。47。53，59，61，67。
71，73，79。83，89，97，101。103。107，
109，113，127，131。137，139，149，151 
*/

void dfs(int temp)  //环当前位置 
{
  int i;
  if(temp==n && prime[way[0]+way[n-1]]==0)//若到达边界且首尾两个数字相加满足素数
  {
    for(i=0;i<n-1;i++)//打印方案
    {
      printf("%d ",way[i]);
    }
    printf("%d\n",way[i]); 
  }
  else
  {
    for(i=2;i<=n;i++)
    {
      if(visit[i]==0 && prime[i+way[temp-1]]==0)  //i没有訪问过且与前一个数相加为素数
      {
        way[temp]=i;
	visit[i]=1;   //表示i訪问过 
	dfs(temp+1);
	visit[i]=0;   //i不满足条件， 之前做的标记要改动 
      }
    }
  }
}

int main()
{
  int count=1;
  memset(visit,0,sizeof(visit));
  //memset(way,0,sizeof(way));
  way[0]=1;
  isprime();
  while(~scanf("%d",&n))
  {
    printf("Case %d:\n",count++);
    dfs(1);
    printf("\n");
  }
  system("pause");
  return 0;
}