题目回顾（HDU-1016）：

Prime Ring Problem

Problem Description

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

源码与解析

#include<iostream>
#include<string.h>
using namespace std;

int judge[41]={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,
	1,0,0,0,0,0,1,0,0,0
};//素数的打表 

int mark[21];//用作标记是否这个数已经使用过
int y[21];//答案数组
int n;

void dfs(int a){
	//如果所有数字确定，那么判断首尾相加是否也是素数，
	//若是，则满足条件，输出
	if(a==n&&judge[y[1]+y[n]]){
		for(int i=1;i<n;i++){
			cout<<y[i]<<" ";
		}
		cout<<y[n]<<endl;
		return;
	}

	//深搜，不断找出未被标记且与此数相加为素数的下一个数
	for(int i=1;i<=n;i++){
		if(mark[i]==0&&judge[y[a]+i]){
			y[a+1]=i;
			mark[i]=1;
			dfs(a+1);
			mark[i]=0;
		}
	}
}

int main(){
	int c=1;
	while(cin>>n){
		cout<<"Case "<<c++<<":"<<endl;
		memset(y,0,sizeof(y));
		memset(mark,0,sizeof(mark));
		y[1]=1;
		mark[1]=1;
		dfs(1);
		cout<<endl;
	}
	return 0;
}

原文地址：https://www.cnblogs.com/orangecyh/p/9873675.html