题目：
Prime Ring Problem
**Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 45061 Accepted Submission(s): 19969
**
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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.
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Input
n (0 < n < 20).\
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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.\
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Sample Input
6
8
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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
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思路：
利用搜索，枚举每一种可能，然后把符合条件的输出
代码：
#include <stdio.h>
#include <string.h>
#define mem(a,b) memset(a,b,sizeof(a))
int prime[100]= {1,1};
int a[50],md[50],n;//a来存放素数环里的数，md是做标记
void sushu()
{
int i,j;
for(i=2; i<=10; i++)
if(prime[i]==0)
for(j=2*i; j<=50; j+=i)
prime[j]=1;//把不是素数的标记为1
}
void dfs(int step)
{
int i,j;
if(step==n+1&&prime[a[n]+a[1]]==0)//结束条件，这个代码是从a[1]开始的
{
for(i=1; i<n; i++)
printf("%d ",a[i]);
printf("%d\n",a[n]);
return;
}
for(i=2; i<=n; i++)
{
if(md[i]==0&&prime[i+a[step-1]]==0)
{
a[step]=i;//给素数环填数
md[i]=1;//标记一下走过的
dfs(step+1);//搜索下一步
md[i]=0;//标记回来
}
}
return;
}
int main()
{
int num=1;
a[1]=1;//从1开始的
sushu();
while(~scanf("%d",&n))
{
mem(md,0);
printf("Case %d:\n",num++);
dfs(2);
printf("\n");
}
return 0;
}
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