ACM 搜索 hdu1016 Prime Ring Problem

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
68
 

Sample Output
Case 1:1 4 3 2 5 61 6 5 2 3 4Case 2:1 2 3 8 5 6 7 41 2 5 8 3 4 7 61 4 7 6 5 8 3 21 6 7 4 3 8 5 2

注意：最后一个数要和1检验是否为素数，这是一个环

#include <stdio.h>int num[21],mark[21],n;int prime_num[12] = {2,3,5,7,11,13,17,19,23,29,31,37};//判断是否是质数，是返回1，不是返回0int is_prime(int a){    for(int i = 0; i < 12;i++)    if(a==prime_num[i])return 1;    return 0;}void print_num(){    for(int i = 1; i < n;i++)    printf("%d ",num[i]);    printf("%d",num[n]);}int dfs(int pre,int post,int flag){    //如果不符合，直接返回    if(!is_prime(pre+post))    return 0;    num[flag] = post;    if(flag==n&&is_prime(post+1))    {        print_num();        printf("\n");        return 1;    }    //使用过了这个数字就标记为0    mark[post] = 0;    for(int i = 2;i<=n;i++)    if(mark[i]!=0 && dfs(post,i,flag+1))break;    //标记位恢复原状    mark[post] = 1;    return 0;}int main(){    int count;    count = 1;    while(scanf("%d",&n)!=EOF)    {        for(int i = 1; i <= n; i++)        mark[i] = i;        num[1] = 1;        printf("Case %d:\n",count++);        if(n==1)printf("1\n");        for(int i = 2;i<=n;i++)        dfs(1,i,2);        printf("\n");    }    return 0;}