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