Prime Ring Problem（回溯法）

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

思路：

回溯法，这个类似于“n皇后”问题。

这个算法用楼梯来形象的描述一下是：当你站在某阶时，判断是否到地面（到地面就结束），不到地面，那么就考虑要下到（题目中限制的)第几阶，并且记录（book），最后归零

实现代码：

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
int n,a[50],book[50];/*这里的a对应输出的数*/
int prime(int x)/*判断素数，是素数返回1*/
{
	for(int i=2;i*i<=x;i++)
	{
		if(x%i==0)return 0;
	}
	return 1;
}
void dfs(int y)
{
	if(y==n&&prime(1+a[n-1])==1)/*判断环里面最后一个数和第一个数1的和是否为素数*/
	{
		printf("1");
		for(int i=1;i<=n-1;i++)
		{
			printf(" %d",a[i]);
		}
		printf("\n");
	}
	for(int i=2;i<=n;i++)
	{
		if(!book[i]&&prime(a[y-1]+i)==1)
		{
			a[y]=i;
			book[i]=1;
			dfs(y+1);
			book[i]=0;/*这里的最后归零很重要！！*/
		}
	} 
}
int main()
{
	int k=0;
	while(~scanf("%d",&n))
	{
		printf("Case %d:\n",++k);
		memset(book,0,sizeof(book));
		a[0]=1;
		dfs(1);
		printf("\n");	/*格式需要*/
	}
	return 0;
}

注意：格式！刚开始阿久把 printf("\n");    /*格式需要*/ 放在了printf("Case %d:\n",++k);上面：写的是if(k>0)printf("\n");/*格式需要*/  提交，Presentation Error 。所以注意格式呀Q.Q

补充：判断素数的函数（更高效）

int is_prime(int x)

{
    for (int i = 2; i*i  <= x; i++)
    if (x % i == 0) return 0;
    return 1;
}