2013
11-26

# 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.

n (0 < n < 20).

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.

6
8

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

#include<stdio.h>
#include<string.h>
int prime[40]={0,1,1,1,0,1,0,1,0,0,0,1,0,1,0,0,0,1,0,1,0,0,0,1,0,0,0,0,0,1,0,1,0,0,0,0,0,1,0,0},n;//素数打表，因为n最大是20，所以只要打到40
int visited[21],a[21];
void dfs(int num)//深搜
{
int i;
if(num==n&&prime[a[num-1]+a[0]])  //满足条件了，就输出来
{
for(i=0;i<num-1;i++)
printf("%d ",a[i]);
printf("%d\n",a[num-1]);
}
else
{
for(i=2;i<=n;i++)
{
if(visited[i]==0)//是否用过了
{
if(prime[i+a[num-1]]) //是否和相邻的加起来是素数
{
visited[i]=-1;//标记了
a[num++]=i;//放进数组
dfs(num); //递归调用
visited[i]=0; //退去标记
num--;
}
}
}
}
}
int main()
{
int num=0;
while(scanf("%d",&n)!=EOF)
{
num++;
printf("Case %d:\n",num);
memset(visited,0,sizeof(visited));
a[0]=1;
dfs(1);
printf("\n");
}
return 0;
}