Z - 送一个dfs喽 深度搜索相加为质数

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

#include<stdio.h>
#include<string.h>
int n;
int a[1000];
int vis[1001];
bool judge(int xx)
{
    int i;
    int flag=1;
    for(i=2; i<xx; i++)
        if(xx%i==0)
        {
            flag=0;
            break;
        }
    if(flag)
        return 1;
    return 0;
}
void dfs(int x,int cnt)
{
    int i;
    if(cnt==n)
    {
        if(judge(x+1))
        {for(i=1; i<=n; i++)
            {
                if(i==n)
                printf("%d\n",a[i]);
            else printf("%d ",a[i]);
            }
        return ;
    }
    else return ;
    }
    for(i=2; i<=n; i++)
    {
        if(!vis[i])
        {
            if(judge(x+i))
            {
                vis[i]=1;
                cnt++;
                a[cnt]=i;
                dfs(i,cnt);
                vis[i]=0;
                cnt--;
            }
        }
    }
    return ;
}
int main()
{
  int ans=0;
    while(~scanf("%d",&n))
    {
        printf("Case %d:\n",++ans);
        memset(vis,0,sizeof(vis));
        vis[1]=1;
        a[1]=1;
        dfs(1,1);
        printf("\n");
    }
    return 0;
}