Prime Ring Problem

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 
    
    

暴力做肯定超时，所以用深搜做。每次判断是否满足素数，如果满足继续搜下去，直到有n个数，同时把前40个数是否是素数保存起来可省不少时间

#include <iostream>
#include <algorithm>
#include <cstring>

using namespace std;

int a[21],n,b[21];//a数组用来存数，b数组用来判断数有没有用过

bool isprime(int t){
	for(int i=2; i*i <= t; ++i){
		if(t%i==0)return false;
	}
	return true;
}
bool ans[41];

void dfs(int k){
	if(k > n) return ;
	if(k==n){
		for(int i=0; i < n-1; ++i) cout << a[i] << " ";
		cout << a[n-1] << endl;
		return ;
	}
	for(int i=2; i <= n; ++i){
		if(b[i]==1) continue;
		if(k+1==n){
			if(ans[i+a[k-1]]&&ans[i+1]){
				a[k]=i;
				b[i]=1;
				dfs(k+1);
				b[i]=0;
			}
		}
		else{
			if(ans[i+a[k-1]]){
				a[k]=i;
				b[i]=1;
				dfs(k+1);
				b[i]=0;
			}
		}
	}
}
int main()
{
	int k=1;
	for(int i=0; i < 41; ++i) ans[i]=isprime(i);
	while(cin >> n){
		cout << "Case " << k << ":" << endl;
		k++;
		memset(a,0,sizeof(a));
		memset(b,0,sizeof(b));
		a[0]=b[0]=1;
		dfs(1);
		cout << endl;
	}
	
}