题目描述：

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.


输入：

n (1 < n < 17).

输出：

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.

样例输入： 
6
8
样例输出：
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
提示：

用printf打印输出。

#include<iostream>
#include<stdio.h>
#include<queue>
using namespace std;

int ans[17];
bool mark[17];
bool check(int a){
    
    if ((a%2 == 0||a%3==0||a%5==0)&&a!=2&&a!=3&&a!=5)//因为输入的n<17，所以相邻数字加和小于32，合数的因子就235 
    return false;
    return true;
}
int n;
void dfs(int cnt){
    if (cnt>1  )
    if (check(ans[cnt-1]+ans[cnt-2])==false)
    return;
    if (cnt==n)
    if (check(ans[cnt-1]+ans[0])==false)
    return;
    if (cnt == n){
        printf("1");
        for (int i=1;i<n;i++){
            printf(" %d",ans[i]);
        } 
        printf("\n");
    }
    for (int i=2;i<=n;i++){
        if (mark[i]==false){
            mark[i]=true;
            ans[cnt]=i;
            //cnt++;
            dfs(cnt+1);
            mark[i]=false;
        }
        
    } 
}

int main (){
    int c=0;
    while (cin>>n){
        c++;
        for (int i=1;i<=n;i++)
        mark[i]=false;
        ans[0]=1;
        mark[1]=true;
        
        printf("Case %d:\n",c);
        dfs(1);
        printf("\n");
    }


    return 0;
}

哎呀呀！他说我超时!!!!找了好久好久好久的bug

后来突然想起之前学长说printf比cout快

然后换了，ac了

刚刚粘题目的时候发现。。提示里面有，我没看见。哭哭

所以说啊，printf好啊，以后尽量用这个吧

这道题也很好啊，

标红是核心，好好看