6 8Sample 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 <iostream>
#include <cstdio>
#include <string>
#include <queue>
#include <stack>
#include <algorithm>
#include <cmath>
#include <list>
#include <cstdlib>
#include <cstring>
using namespace std;
int n,a[30],vis[30];//vis是标记数组
int prime[40]={0,1,1,1,0,1,0,1,0,0,0,1,0,1,0,0,0,1,0,1,0,0,0,1,0,0,0,0,0,1,0,1,0,0,0,0,0,1,0,0};//1-40素数表，素数用1标记
void dfs(int num)
{
if(num==n&&prime[a[num-1]+a[0]])
{
for(int i=0;i<num-1;i++)
{
printf("%d ",a[i]);
}
printf("%d\n",a[num-1]);
}
else
{
for(int i=2;i<=n;i++)
{
if(vis[i]==0&&prime[i+a[num-1]])//判断i的标记和素数表
{
vis[i]=1;
a[num++]=i;
dfs(num);
vis[i]=0;//回溯标记和num
num--;
}
}
}
}
int main()
{
int num;
num=0;
while(cin>>n)
{
memset(vis,0,sizeof(vis));
num++;
printf("Case %d:\n",num);
a[0]=1;
dfs(1);
printf("\n");
}
return 0;
}
参考：https://blog.csdn.net/feng_zhiyu/article/details/75576867
原文：https://www.cnblogs.com/FantasticDoubleFish/p/14386396.html