解体心得：
1、一个回溯法，可以参考八皇后问题。
2、题目要求按照字典序输出，其实在按照回溯法得到的答案是很正常的字典序。不用去特意排序。
3、输出有个坑，就是在输出一串的最后不能有空格，不然要PE，很尴尬。
题目：
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 3 8 5 2
Source
Asia 1993 8 5 2
Source
Asia 19 Shanghai (Mainland China)
#include<stdio.h>
#include<cstring>
int n,num[21],sum,l,sum1;
bool prim[50];
bool bnum[21];
int circle[21];
int z = 1;
void dfs(int sum)
{
if(sum == n)
{
if(!prim[circle[1]+circle[sum]])//判断第一个和最后一个相加是否是素数
return;
for(int j=1;j<=n;j++)
{
if(j!=n)
printf("%d ",circle[j]);//最后一个数字后面不可以加空格，不然要PE，很尴尬；
else
printf("%d",circle[j]);
}
printf("\n");
return;
}
for(int i=2;i<=n;i++)
{
if(!bnum[i])
continue;
if(prim[i+circle[sum]])
{
circle[sum+1] = i;
bnum[i] = false;
dfs(sum + 1);
bnum[i] = true;
}
}
return;
}
int main()
{
//先把50以内的素数手写出来
memset(prim,false,sizeof(prim));
prim[2] = true;
prim[3] = true;
prim[5] = true;
prim[7] = true;
prim[11] = true;
prim[13] = true;
prim[17] = true;
prim[19] = true;
prim[23] = true;
prim[29] = true;
prim[31] = true;
prim[37] = true;
prim[41] = true;
prim[43] = true;
prim[47] = true;
while(scanf("%d",&n)!=EOF)
{
//特判几个很大但是没有任何结果的数，直接输出，节省时间
if(n == 19 || n == 15 || n == 11 || n == 17 || n == 13)
{
printf("Case %d:\n",z++);
printf("\n");
continue;
}
printf("Case %d:\n",z++);
sum1 = 0;
l = 1;
sum = 1;
memset(circle,0,sizeof(circle));
memset(bnum,true,sizeof(bnum));
bnum[1] = false;
circle[1]=1;
dfs(1);
printf("\n");
}
}