dfs hdu 1016 Prime Ring Problem 素数环

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

题意：
给一个n求n的全排列中相邻两个和是素数包括第一个和最后一个的排列；
思路：
由于n最大20所以可以打一个素数表， if(num==n&&prime[a[n-1]+a[0]])搜索结束条件，其他的看注释；

代码：

#include<cstdio>#include<iostream>#include<algorithm>#include<cstring>#include<cmath>using namespace std;int prime[40]= {0,0,1,1,0,1,0,1,0,0,0,1,0,1,0,0,0,1,0,1,0,0,0,1,0,0,0,0,0,1,0,1,0,0,0,0,0,1};int n;int a[100],vis[100];void dfs(int num){    if(num==n&&prime[a[n-1]+a[0]])    {        for(int i=0; i<n-1; i++)        {            printf("%d ",a[i]);        }        printf("%d\n",a[n-1]);    }    else    {        for(int i=2; i<=n; i++)        {            if(!vis[i]&&prime[i+a[num-1]])            {                vis[i]=1;//标记；                a[num]=i;//num此时等于1//由于第一个数是1，所以1已经被用过了，得从2开始往数组a里面存；                dfs(num+1);                vis[i]=0;   //回溯；            }        }    }}int main(){    int t=0;    while(~scanf("%d",&n))    {        t++; printf("Case %d:\n",t);        memset(vis,0,sizeof(vis));        a[0]=1;        dfs(1);        printf("\n");    }}