題目：

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

輸入：

n (1 < n < 17).

輸出：

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.

樣例輸入：

6
8

樣例輸出：

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

解題思路：

dfs和回溯的又一好例子，注意return的用法。

# include<stdio.h>
# define N 17
int prime[]={2,3,5,7,11,13,17,19,23,29,31,37};//這題重點並不在於求解素數本身

int n;
bool hash[N];
int ans[N];

bool check(int x)
{
	for(int i=0;i<12;i++)
	{
		if(x==prime[i])
			return true;
	}
	return false;
}

void final_put()
{
	if(check(ans[n]+ans[1])==false)//不往下執行了
		return;

	else
	{
		for(int i=1;i<=n;i++)
		{
			if(i==1)
				printf("%d",ans[i]);
			else
				printf(" %d",ans[i]);
		}
		printf("\n");
	}
}

void dfs(int num)
{
    if(num>=2)
	{
        if(check(ans[num]+ans[num-1])==false)//回退上一級
			return;
	}
	
	if(num==n)//設成全域性變數
	{
		final_put();
		return;//回退上一級
	}
    
	for(int i=2;i<=n;i++)//遍歷
	{
		if(hash[i]==false)
		{
			hash[i]=true;
			ans[num+1]=i;
			dfs(num+1);//呼叫

			hash[i]=false;//全部呼叫完成後，返回時賦值。
		}
	}
}

int main()
{
	int i;
	int mycase=0;
	while(scanf("%d",&n)!=EOF)
	{
        mycase++;
		printf("Case %d:\n",mycase);
        //初始化
		for(i=1;i<N;i++)
		{
			hash[i]=false;
			ans[i]=-1;
		}

		hash[1]=true;
		ans[1]=1;

		dfs(1);
		printf("\n");
	}
	return 0;
}